\insertitre{A. Proof of the renormalization lemma.}{}
\section*{ A. Proof of the renormalization lemma.}


\noindent {\it (i) For linear transforms $B$ and $C$ and any $t$ one has
the semigroup property $t'(t,BC)=t'(t'(t,B),C)$.}
 \rm Indeed, we have $BCT_{t'(t,BC)}=T_tBC=BT_{t'(t,B)}C=BCT_{t'(t'(t,B),C)}$.
The map which associates $T_t$ with t being one to one, this implies the
stated relation.

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\noindent {\it (ii) $t'(t,B)$  is increasing with respect to $t$.}

 \rm Let us prove that $t'(t,A)$ is one to one with respect to $t$ 
for any $A$. Indeed, if not, there would be some $A$ and some $(s,t)$
such that $t'(t,A)=t'(s,A)$. Thus $T_tA=AT_{t'(t,A)}=AT_{t'(s,A)}=T_sA$ and 
 therefore $t=s$ because $T_t$ is one to one.

\noindent Since $t'(t,A)$ is non negative (by definition), one to one,
continuous with respect to $t$ and $t'(0,A)=0$ we can deduce that it is 
increasing with respect to $t$.

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\noindent {\it Moreover if $B$ is orthogonal then  $t'(t,B)=t$.}

 \rm Let $R$ be orthogonal. Then iterating the formula of (i) we have:
 $$t'(t'(t'(t'(...t'(t,R)...,R),R),R)=t'(t,R^n)$$
Remark that there is a subsequence of $R^n$ tending to $Id$.
Indeed, there is a subsequence $R^{n_k}$ which converges
to $H$ orthogonal because the orthogonal group is compact. Therefore, the
subsequence $R^{n_{k+1}-n_k}$ converges to $Id$ .
Since there exists a subsequence of $R^n$ tending to $Id$ and since 
$t'$ is continuous we have for this subsequence $lim \; t'(t,R^n)=t'(t,Id)=t$.
 Assume by contradiction that $t'(t,R)=t"$ with $t"<t$ then $t'(t'(t,R),R)=t'(t",R)\le t'(t,R)=t"$
and by recursion, $t'(t,R^n)=t'(t'(t'(t'(...t'(t,R)...,R),R),R)\le t"<t$.
This is a contradiction. Thus $t'(t,R)\ge t$. We prove the converse 
inequality in the same way and we obtain $t'(t,R)=t$.
   

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\noindent (iii) For any orthogonal R, we have proved that $t'(t,R)=t$.
Now, we note that any linear transform $B$ of $\Re ^2$ can be obtained
as a product of orthogonal transforms and of linear transforms of the kind $A(\lambda)$:
$(x,y) \rightarrow (\lambda x,y)$ where $\lambda$ is non negative.
We only need to make a singular value decomposition of $B$:
  $B=R_1DR_2$, where $R_1$ and $R_2$ are both orthogonal transforms
and $D$ is a transform of the kind $(x,y) \rightarrow ({\lambda}_1 x,{\lambda}_2 y)$ where $\lambda_i$ are non negative. 
Now, it is clear that $D$ can be decomposed as $D=A({\lambda}_1)RA({\lambda}_2)R^{-1}$ where
R is the orthogonal transform: $(x,y) \rightarrow (-y,x)$.
Using (i), $t'(t,R_i)=t$ for any orthogonal transform $R$ , the singular value decomposition and 
$A({\lambda}_1)A({\lambda}_2)=A({\lambda}_1{\lambda}_2)$, we obtain
 $$t'(t,B)=t'(t,{\lambda}_1{\lambda}_2)=t'(t,\vert detB{\vert}^{1/2})$$   

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\noindent (iv) Using (i) and (ii), we have $t'(t,\lambda \mu)=t'(t'(t,\mu)\lambda)$
 for any positive $\lambda$ and $\mu$. Differentiating this relation with respect
to $\mu$ at $\mu = 1$ yields
$$(0) \; \lambda \frac{\partial t'}{\partial \lambda}(t,\lambda)=\frac{\partial t'}{\partial \lambda}(t,1)\frac{\partial t'}{\partial t}(t,\lambda)$$   

Choose $h$ such that $\phi h'=h$ and set $t'(t,\lambda)=G(t,h(t)\lambda)$, where
$$h(t)=exp(\int_1^t ds/\phi (s))$$
Then the preceding relation (0) yields $\frac{\partial G}{\partial x}(x,y)=0$.
Thus $G(x,y)=\beta (y)$ for some differentiable nondecreasing function $\beta$.
We obtain that $t'(t,\lambda)=\beta (h(t)\lambda )$. Returning to the 
definition of $\phi (t)$, we have
$\phi (t)=\frac{\partial t'}{\partial \lambda}(t,1)=\frac{\partial \beta(h(t)\lambda )}{\partial \lambda}(t,1)$
 and $\phi (t)=h(t)\beta '(\lambda h(t))=\phi (t)h'(t)\beta '(h(t)\lambda)$
 thus the derivative of $\beta (h(t))$ is $1$ and integrating
this last relation between $0$ and $t$ yields $\beta (h(t))=t+\beta (h(0))$.
Using the fact that $t'(0,\lambda )=0$ (which derives from injectivity
of the $T_t$), we obtain $\beta (h(0))=0$ and therefore $t'(t,\lambda ) = h^{-1}(\lambda h(t))$.   
To finish the proof, we set $S_t=T_{h^{-1}(t)}$ and we prove that the
affine invariance is true for $S_t$ with $t'(t,\lambda)=\lambda t$.
$ S_tB=T_{h^{-1}(t)}B=BT_{t'(h^{-1}(t),\lambda )}=BT_{h^{-1}(\lambda h(h^{-1}(t)))}=$
$BT_{h^{-1}(\lambda t)}=BS_{\lambda t}$.



