Introduction
============

An integrable Hamiltonian system is given in the action-angle coordinates $(I, \theta)\in {\mathbb{T}}^d\times {\mathbb{R}}^d$ by a Hamiltonian $H_0(I)$ depending only on the actions. It generates the flow $$\label{int}
\dot I=0, \ \ \dot\theta=\omega(I)$$ where $\omega=\partial_I H_0$. The orbit $\theta(t)=\omega(I(0))t$ is quasiperiodic and spans the torus $I=I(0)$ for almost all $I(0)$ in a neighbourhood of a point $I$ where the Hessian $D^2H_0$ is nondegenerate. The KAM theorem constructs invariant tori in such a neighbourhood when $H_0$ is perturbed by a Hamiltonian $\lambda H_1(I,\theta)$ suitably small in the analytic ([@kol], [@ar]) or $C^k$ [@mos] category. These solutions are found by a Newton iteration that constructs coordinates where the flow takes the simple form .

An alternative approach in the analytic case is a perturbation expansion of the solutions in powers of $\lambda$. The resulting Lindstedt series which converges as a consequence of the KAM theorem [@kol] was proven to do so by directly bounding the coefficients much later by Eliasson [@eli] who showed the existence of huge cancellations in the terms in the series. Subsequently Eliasson’s work was simplified and extended by Gallavotti [@g1; @g2; @g3], by Chierchia and Falcolini [@gf] and by Bonetto, Gentile and Mastropitero [@bgm]. In particular Gallavotti [*et. al.*]{} revealed the analogy of Lindstedt series to perturbative expansions in quantum field theory (QFT) and used ideas, in particular the renormalization group (RG), that they had developed for the latter to the former.

These papers in turn were an inspiration to the study [@bgk] which developed a new iterative scheme, inspired by the RG. In this approach the small denominators that plague the Lindstedt series are studied scale by scale by introducing renormalized problems for the unresolved scales. In the iterative scheme the renormalized problems become simpler as the scale of the denominators is reduced. This fact was shown to be a consequence of symmetries in the problem which fact provides an explanation for the cancellations that Eliasson uncovered in his proof. The renormalization group iteration was subsequently extended to PDE’s [@bks], to the homoclinic splitting problem [@mikko] and also to the case of a $C^k$ perturbation [@emil] where the Lindstedt series is not convergent.

In this paper we return to the analytic KAM theorem and give a proof of it using a RG where the scale is changed continously. The renormalized problems are given as solutions of a simple evolution PDE with quadratic nonlinearity. This equation is well known in QFT where it goes under the name Polchinski equation [@pol]. Although conceptually this proof is just an infinitesimal version of the one given in [@bgk] we feel it is worth writing down for two reasons.

First, the KAM theorem is reduced to a very simple Picard iteration with elementary estimates.

Secondly we wish to bring into attention the Polchinski equation in this context. It is well known that as the size $\lambda$ of the perturbation is increased the torus with a given frequency $\omega$ disappears. For certain “scale invariant” $\omega$ (e.g. in $d=2$ for $\omega=(1,\gamma)$ with $\gamma$ a “noble” irrational) the solution at the critical $\lambda$ is believed to have “universal” scaling properties [@kada; @shenker]. Such universality presumably would be related to fixed points of the RG equation (composed with a scaling). It would be very interesting to see if the RG equation possesses such scaling solutions. The simplicity of the equation means that this hope might not be entirely unrealistic.

The RG Equation
===============

We will give the proof of KAM theorem in the special case where the perturbation $ H_1$ is a function of the angles $\theta$ only. This case possesses all the difficulties of the general one (for the RG formalism in the general case see [@bgk]). Thus, given a real analytic function $v:{\mathbb{T}}^d\to {\mathbb{R}}$ where ${\mathbb{T}}$ is the circle ${\mathbb{R}}/2\pi{\mathbb{Z}}$, we look for quasiperiodic solutions to the equation $$\label{newton}
\ddot \theta=-\lambda\partial v(\theta)$$ where $\lambda$ is a perturbation parameter. The solutions are constructed by by fixing a vector $\omega\in{\mathbb{R}}^d$ whose components are independent over the integers and looking for a function $X:{\mathbb{T}}^d\to {\mathbb{R}}^d$ s.t. $$\theta(t)=\omega t+X(\omega t)$$ solves . This follows, if $X$ solves the equation $$\label{main0}-(\omega\cdot\partial )^2X= U(X)$$ where $$\label{main1}
U(\theta,X) = \lambda \partial v(\theta + X(\theta)) .$$ Note that if $X$ solves so does for all $\beta\in{\mathbb{R}}^d$ $$\label{transla}
X_\beta(\theta)=X(\theta+\beta)+\beta.$$ We work mostly in Fourier space using lower case letters for the Fourier transform $$x(q)=\int_{{\mathbb{T}}^d}e^{-iq\theta}X(\theta)
{\nonumber}$$ for $q\in{\mathbb{Z}}{^d}$. Eq. can be written as fixed point equation $$\label{main}
x(q)= (\omega q)^{-2} u(q,x)\ \ q\neq 0,\ \ \ u(0,x)=0$$ which is nontrivial since the multiplier $(\omega q)^{-2}$ is unbounded because the denominator can get arbitrary small.

For the RG we cut out small denominators $ \omega q$ in and study how the problem changes when the cutoff is changed. Let $\chi:{\mathbb{R}}^+\to [0,1]$ be a smooth monotone function with $\chi=0$ on $[0,1]$ and $\chi=1$ on $[2,\infty)$ and let $$\eta(t) = te^{\alpha t}$$ where $\alpha>0$ is chosen later. Define $$\label{Gt }
\gamma_t(q)=\chi(\eta(t) |\omega q|)(\omega q)^{-2}.$$ Thus $\gamma_t$ has the small denominators $|\omega q|\leq \eta(t)^{-1}$ cut off.

We will look for a solution to the cutoff problem $$\label{xt}
x(t)= \gamma_t u(x(t))$$ by studying a bit more general one. Suppose we can find a functional $f(t,y)$ defined in a suitable space of maps $y:{\mathbb{T}}^d\to{\mathbb{R}}^d$ and satisfying $$\label{ft}
f(t,y) = y + \gamma_t u(f(t,y)).$$ Then $x(t)=f(t,0)$. Define also $$\label{wt}
w(t,y) = u(f(t,y)).$$ Then, differentiating in $t$ and $y$ we have $(1-\gamma Du(f))\dot f=\dot\gamma w$ and $(1-\gamma Du(f))Df=1$ i.e. we should look for $f$ as a solution to the differential equation $$\label{Feq}
\dot f(t,y) = Df(t,y)\dot\gamma_t w(t,y) \, , \quad f(0,y) = y$$ where the initial condition is due to $\gamma_0=0$. For $w$ we get from $\dot w=Du(f)\dot f=Du(f)Df\dot\gamma \ w$ i.e. we look for the solution to $$\label{EqW}
\dot w(t,y) = Dw(t,y)\dot\gamma_t w(t,y) \, , \quad w(0, y) = u(y).$$ Our strategy then is to solve the RG equations and in a suitable space and construct the solution of as the limit $$\label{Final}
x = \lim_{t\rightarrow \infty} f(t,0).$$ We need the usual Diophantine property: for some $a,\nu>0$ and all $q\neq 0$ $$\label{dio}
|\omega q|\geq a|q|^{-\nu}.$$ Then we prove

Let $v$ be real analytic. Then the equation has a unique, up to translations , real analytic solution analytic in $\lambda$ in some neighbourhood of the origin.

Identities
==========

The flow possesses two symmetries, exhibited in [@bgk], that are crucial for the analysis. Before defining them let us write in the Fourier variables $$\label{EqW1}
\dot w(t,q,y) = \sum_{r\in{\mathbb{Z}}{^d}}Dw(t,q,r,y)\dot\gamma_t (\omega r)w(t,r,y)$$ where $Dw(t,q,r,y)$ is a $d\times d$ matrix and setting $\kappa=\omega r$ $$\label{gamma}
\dot\gamma_t = \kappa^{-2} \frac{d}{dt} \chi(\eta(t) |\kappa|)$$ $w$ and $f$ will be given by their Taylor series $$\label{w}
w(t,q,y)= \sum_{n=0}^{\infty} \sum_{{\boldsymbol{p}}\in {\mathbb{Z}}^d \times \cdots \times {\mathbb{Z}}^d} w_n(t,q, p_1, \ldots , p_n) (y(p_1), \ldots , y(p_n)).$$ The kernels $w_n$ are linear maps $w_n\in{\mathcal L}(({\mathbb{R}}^d)^{n}, {\mathbb{R}}^d)$, symmetric in $p_1, \ldots p_n$; for $f$ we have a similar expansion. The equation becomes then a set of equations for the kernels: $$\label{kernels}
\dot w_n(t, q, {\boldsymbol{p}}) = \sum_{k=1}^{n+1} \sum_{r\in {\mathbb{Z}}^d} k w_k(t,q,r,{\boldsymbol{p}}')w_{n+1-k}(t,r, {\boldsymbol{p}}'') \dot\gamma_t(\omega r)   \equiv
v_n(t, q, {\boldsymbol{p}})$$ with ${\boldsymbol{p}}= ({\boldsymbol{p}}' , {\boldsymbol{p}}'')$. Eq. should be symmetrized in the ${\boldsymbol{p}}$ but we will keep this implicit in the notation.

 ${\mathbb{Z}}^d$ action
------------------------

To define this note first that at $t=0$ we get from after some calculation $$\label{un}
u_n(q, {\boldsymbol{p}}) ={\lambda v(r)\over n!} (ir)^{\otimes^{ n+1}},
\quad
r=q-\sum p_i$$ where from now on we identify ${\mathcal L}(({\mathbb{R}}^{d})^n, {\mathbb{R}}^d)$ with $({\mathbb{R}}^d)^{ n+1}$. Hence, trivially, $$\label{shift0}
u_n(q + p, p_1+p,{\boldsymbol{p}}') =  u_n(q , p_1,{\boldsymbol{p}}')$$ for $p \in {\mathbb{Z}}^d$. This symmetry will, for $t > 0$ take a peculiar form. Let $$\pi(t,q,q',y)=Dw(t,q,q',y).$$ Then can be paraphrased as $\tau_p\pi(0,q,q',y)\equiv \pi(0,q+p,q'+p,y)=\pi(0,q,q',y)$. On the other hand, yields $$\nonumber
\dot \pi(t,q,q',y) = \sum_{r}(D\pi(t,q,q',r,y)w(t,r,y) +\pi(t,q,r,y) \pi(t,r,q',y) )\dot\gamma_t (\omega r).$$ The translated function satisfies upon shifting $r$ by $p$ in the second term and denoting $\omega p=\kappa$ \[pi\] \_p(t,q,q’,y)& =& \_[r]{}D\_p(t,q,q’,r,y)w(t,r,y)\_t (r)\
&+&\_[r]{}\_p(t,q,r,y) \_p(t,r,q’,y)\_t (r+) . Suppose now we can find a one parameter family of functions $\pi(t,q,q',y,\kappa)$, $\kappa\in{\mathbb{R}}$ solving eq. with $\tau_p \pi$ replaced by $\pi(\cdot,\kappa)$. with initial condition $\pi|_{t=0}=Du$ independently on $\kappa$. Uniqueness of solutions will then yield $$\label{tra}
Dw(t,q+p,q'+p,y)=\pi(t,q,q',y,\omega p)$$ which will be used in the second symmetry. The equation for $\pi(\cdot,\kappa)$ can be written in terms of kernels $w_n(q,q',{\boldsymbol{p}},\kappa)$, ${\boldsymbol{p}}\in({\mathbb{R}}^d)^{(n-1)}$, symmetric ${\boldsymbol{p}}$, which at $n=0$ or $\kappa=0$ coincide with $w_n(q,q',{\boldsymbol{p}})$. For $n>0$ eq. is modified to $$\label{kernels1}
\dot w_n(t,q,q', {\boldsymbol{p}},\kappa)=v_n(t,q,q', {\boldsymbol{p}},\kappa)$$ with \[vn\] v\_n(t,q,q’, ,)&=& \_[k+l=n+1]{} \_[r]{} \[(k-1) w\_k(t,q,q’,r,’,)w\_[l]{}(t,r, ”)) \_t(r)\
&&+ l w\_k(t,q,r,’,)w\_[l]{}(t,r, q’,”,)) \_t(r+)\] where again symmetrization in ${\boldsymbol{p}}$ is implicit. Eq. obviously coincides with eq. at $\kappa=0$, the two positions of $q'$ giving rise to symmetrization in $(q',{\boldsymbol{p}})$.

The ${\mathbb{Z}}^d$ action is then given by $$\label{tr1}
w_n(t,q+r, q'+r,{\boldsymbol{p}}, \kappa) = w_n(t,q,q', {\boldsymbol{p}},\kappa + \omega r)$$ and given $w$ satisfying the vector field $v$ in satisfies it too. Eq finds its use in $$\label{use}
v_1(t,q,q, 0) = v_1(t,0,0, \omega \cdot q) .$$

 $T^d$-action
-------------

Consider now $\beta \in {\mathbb{T}}^d$ and define the action $W \mapsto \tau_{\beta}W$ where $$\label{action}
\tau_{\beta}W (Y)(\theta) = W(Y_{\beta})(\theta - \beta) \quad \text{with} \quad Y_{\beta}(\theta)=  Y(\theta + \beta) + \beta$$ From we have for $t=0$ $
\tau_{\beta} U = U$. Moreover the vector field on the RHS of is invariant under $\tau_\beta$. Indeed, we have $$(D\tau_\beta W)(\theta,\theta',Y)=DW(\theta-\beta
,\theta'-\beta,Y_\beta)$$ so the invariance follows from the fact that the kernel of the operator $\dot\gamma$ in $\theta$ space $\dot\Gamma(\theta-\theta')$ commutes with translations. Thus we will look for solutions satisfying $
\tau_{\beta} w = w$. Infinitesimally $\partial_\beta \tau_{\beta} w=0$ gives the so called *Ward identity*: $$\label{Ward0}
a\cdot\partial_{\theta} W(\theta,Y)= \int_{{\mathbb{T}}^d} 
DW(\theta,\theta',Y)(a+a\cdot\partial_{\theta} Y(\theta'))
 d\theta'$$ for any $a\in{\mathbb{R}}^d$. In Fourier space this reads $$\label{Ward00}
ia\cdot qw(q,y)= \sum_{q'}
Dw(q,q',y)(a\delta_{q'0}+ia\cdot q' y(q'))$$ or, in terms of the kernels $$\label{Wardn}
ia\cdot qw_n(q,{\boldsymbol{p}})= (n+1)w_{n+1}(q,0,{\boldsymbol{p}})a+i\sum_i a\cdot p_i
w_n(q,{\boldsymbol{p}})$$ The $n=0$ relation will play a crucial role later: $$\label{Ward2}
i w_0(q) \otimes q= w_1(q,0).$$ Furthermore, using Eq. together with we get $$\label{dotWard}
\dot w_0(0) = \sum_{r} w_1(0,r)\dot\gamma(r)w_0(r) = -i \sum_r r \otimes w_0(-r)\dot\gamma(r)w_0(r) = 0$$ since the summand is antisymmetric in $r$. Thus, since $w_0(0)|_{t=0} = u_0(0) = \int \lambda \partial_{\theta} v(\theta) d\theta = 0$, Eq. implies $$\label{Ward1}
w_0(0) = 0.$$ To see the consequences of eqs. and note that eq. implies $$\label{note}
u_1(q,q',y) = u_1(-q',-q,y)^T.$$ It is readily checked $\tilde w_m(q,q',{\boldsymbol{p}},\kappa)\equiv w_m(-q',-q,{\boldsymbol{p}},-\kappa)^T$ satisfies equation as well. Then, given uniqueness of solutions to eq. we conclude $$\label{note1}
w_n(q,q',{\boldsymbol{p}},\kappa) = w_n(-q',-q,{\boldsymbol{p}},-\kappa)^T.$$ Consider in particular eq. for $n=1$ and $q=q'=0$. By $w_2(t,0,0,{\boldsymbol{p}},\kappa)$ is even in $\kappa$. Hence $$\label{fin}
\dot w_1(0,0,\kappa)=
\sum_{r} w_1(0,r,\kappa)w_{1}(0, -r,-\kappa))^T \dot\gamma_t(\omega r+\kappa)+{\rm even}$$ where was used. Since $\dot\gamma_t$ is even, we deduce that the RHS is even in $\kappa$ and therefore $w_1(0,0,\kappa)$ is also even. We will show below that $w_n$ are $C^2$ in $\kappa$. Hence $$\label{w3}
\partial_\kappa w_1(0,0,0)=0.$$

Estimates
=========

We describe now the functional space where is solved. Consider first $u$ given by . By assumption, $v$ is analytic in $|\Im\theta_i|\leq 2b$ for some $ b>0$. Then eq. implies $$\label{u}
|u_n| \leq  C|\lambda|R^{n}e^{-b|r|}$$ for some $R$ depending on $b$. Given ${\boldsymbol{q}}\in ({\mathbb{Z}}^{d})^n$ and $\kappa \in {\mathbb{R}}$, let $w_n({\boldsymbol{q}}, \kappa)\in ({\mathbb{R}}^d)^{\otimes^{n+1}}$ be symmetric under simultaneous exchanges of $q_i$ and $q_j$ and the corresponding ${\mathbb{R}}^d$’s for $i,j\geq 3$, for $\kappa=0$ for $i,j\geq 2$ and satisfying , and . Let $$\label{Lambda}
\Lambda_t = \{ q \in {\mathbb{Z}}^d \, | \, \eta(t) |\omega \cdot q| \leq 3 \}$$ and, for $\beta>0$, $$\label{beta}
\beta_t={{_1\over^2}}(1+ \frac{1}{1+t})\beta$$ Define $$\label{normt}
\Vert w_n \Vert_t = \sup_{{\boldsymbol{p}}\in \Lambda_t^n} \sum_{q \in \Lambda_t} e^{\beta_t |q- \sum p_i|}|w_n(q,{\boldsymbol{p}}, \cdot)|_t$$ where $|\cdot|_t$ is the following $\mathcal{C}^2$ norm: $$\label{normC2}
|f|_t = \sup_{\eta(t)|\kappa| \leq 1} \sum_{i=0}^2 \eta(t)^{-i}|\partial_i f(\kappa)|$$ and $|\cdot|$ is the standard norm in $({\mathbb{R}}^d)^{\otimes^{n+1}}$.

Let $w(t)= \{w_n(t)\}_{n \in {\mathbb{N}}}$ and define $$\label{norm}
\Vert w \Vert = \sup_{t \geq 0} \left( e^{2t} \Vert w_0(t) \Vert_t + \eta(t)^2\Vert w_1(t) \Vert_t + \sup_{n > 1}n^2 \rho^ne^{\left(\frac{3}{2}-n\right)t}\Vert w_n(t) \Vert_t \right).$$ We have then

\[mainprop\] The equation has a unique solution analytic in $\lambda$ in some ball around zero and satisfying $$\label{bound0}
\Vert w \Vert \leq C(v) |\lambda|$$ where the constant $C$ depends on $v$.

We write as a fixed point equation $$\begin{aligned}
\label{integral}
w(t)=u+\int_0^t v(s)ds\equiv u+\Phi(t,w)\end{aligned}$$ where $v_n$ is given by for $n=0$ and by for $n>0$.

We start by showing $$\label{unorm}
\Vert u \Vert\leq C\lambda.$$ We use the bound and take $\beta<b$ to get $$\label{unormn}
\Vert u_n \Vert_0\leq CR^{n}|\lambda|.$$ For $n=0$ we have $$\label{u0}
\sum_{q \in \Lambda_t} |u_0(q)|e^{\beta_t |q|}  \leq \sup_{0 \neq q \in \Lambda_t}e^{-(\beta - \beta_t)|q|} \Vert u_0 \Vert_0$$ where $q \neq 0$ since by $u_0(0)= 0$. By the Diophantine condition eq. $|\omega \cdot q| \eta(t) < 3$ implies $|q| \geq \left( \frac{a}{3}\eta(t) \right)^{\frac{1}{\nu}}$. Since $\beta - \beta_t = \frac{1}{2} \beta \frac{t}{1+t}$ the sup is superexponential in $t$, hence $$\label{u0ok}
\Vert u_0 \Vert_t \leq C |\lambda|e^{-2t}.$$ For $n=1$, we get from $u_1(q,q) = 0$, so $$\label{u1}
\sum_{q \in \Lambda_t} |u_1(q,p)|e^{\beta_t |q-p|} \leq \sup_{\substack{q,p \in \Lambda_t\\q \neq p}} e^{-(\beta - \beta_t)|q-p|}\Vert u_1 \Vert_0$$ and since $\eta(t) |\omega \cdot (q-p)| \leq 6$ we get as in $$\label{u1ok}
\Vert u_1\Vert_t \leq CR |\lambda| \eta(t)^{-2}.$$ Finally for $u_n$, $n > 1$ we just use $\beta_t \leq \beta_0$ to conclude $$\label{unok}
\sum_{q \in {\mathbb{Z}}^d} |u_n (q,{\boldsymbol{p}})| e^{\beta_t |q-{\boldsymbol{p}}|} \leq \Vert u_n\Vert_0
\leq CR^ne^{(n- \frac{3}{2})t} |\lambda|\leq Cn^{-2}\rho^{-n}e^{(n- \frac{3}{2})t} |\lambda|$$ provided we take $\rho R<1$. Then Eq. follows from , , and .

Thus it suffices to prove that $\Phi$ is a contraction from the ball $B_u(R|\lambda|)$ to $B_0(R|\lambda|)$ for a suitable $R$. We shall treat separately $\Phi_n$ for $n=0, n=1$ and $n \geq 2$.

$n=0$. From we have $$\label{v0}
v_0(s,q)=\sum_{r \in \Lambda_s} w_1(s,q,r)
w_0(s,r)\dot\gamma_s(\omega r)$$ where we may restrict $r\in\Lambda_s$ since by $\dot\gamma_s(x)$ is supported in $\eta(s)|x| \in [1,2]$. Using $e^{\beta_t |q|} \leq e^{\beta_t|q-r| + \beta_t|r|}$ and $\beta_t\leq\beta_s$ together with the obvious $$\label{obv}
|\cdot |_t \leq |\cdot |_s, \ \ |AB|_s \leq c|A|_s|B|_s$$ we get $$\sum_{q \in \Lambda_t} \!\!e^{\beta_t |q|}|\Phi_0(t,q)|_t 
\leq C \sum_{q \in \Lambda_t} \!  \int_0^t\sum_{r \in \Lambda_s}
 \!\!e^{\beta_s|q-r|}|w_1(s,q,r)|_s e^{\beta_s|r|}|w_0(s,r)|_s |\dot\gamma_s|_s e^{-(\beta_s - \beta_t)|q|}ds.
\label{phi0}$$ From we infer $$\label{gammabound}
|\dot\gamma_s|_s \leq C \eta(s)^2$$ so $$\label{so}
\Vert \Phi_0(t)\Vert_t \leq C \Vert w \Vert^2 \int_0^t e^{-2s} \inf_{0 \neq q \in \Lambda_t} e^{-(\beta_s-\beta_t)|q|}ds.$$ Since $\beta_s-\beta_t\geq{{_1\over^2}}\beta(1+t)^{-2}(t-s)$ the Diophantine property implies $$\label{dp}
e^{-(\beta_s - \beta_t)|q|} \leq Ce^{-(t-s)a(t)}$$ where $a(t) \rightarrow \infty$ as $t \rightarrow \infty$. Hence $$\label{phi0ok}
\Vert \Phi_0(t)\Vert_t \leq C e^{-2t} \Vert w \Vert^2.$$

$n=1$. Write $v_1=v_a+v_b$ with \[va\] v\_a(s,q,q’,) &=& \_[r\_s]{} w\_1 (s,q,r) w\_1(s,r,q’) \_s(r + )\
v\_b(s,q,q’,) &=& \_[r\_s]{} w\_2(s,q,r,q’,) w\_0(s,r) \_s(r) \[vb\] and $\Phi_1$ similarily. We could restrict $r\in \Lambda_s$ also in since $\eta(t)|\kappa| \leq 1$ and so $\eta(s) |\omega  r| \leq 2 + \eta(s)|\kappa| \leq 3  $ because $\eta(t)$ is increasing in $t$.

We consider first the case $q \neq q'$ and proceed as in the $n=0$ case. For $v_a$ we insert $${\nonumber}e^{\beta_t|q-q'|} \leq e^{\beta_s \left[ |q-r| + |r-q'|  \right]} e^{-(\beta_s - \beta_t)|q-q'|}$$ and use , , and to get $$\label{insert2}
\sum_{q \neq q'} |e^{\beta_t|q-q'|} \Phi_{a}(t,q,q')|_t \leq C \Vert w \Vert^2 \int_0^t \eta(s)^{-2}
 e^{-(t-s)a(t)}ds \leq C \eta(t)^{-2} \Vert w \Vert^2$$ Similarily, for $\Phi_{b}$ we use $${\nonumber}e^{\beta_t |q-q'|} \leq C e^{\beta_s (|q-r-q'| + |r|)} 
e^{-(\beta_s - \beta_t)|q-q'|}$$ to get $$\begin{aligned}
\label{phi12}
\sum_{q \neq q'}e^{\beta_t |q-q'|}  |\Phi_{b}(t,q,q')|_t  \leq C \Vert w \Vert^2  \int_0^t e^{s/2 - 2s} \eta(s)^2 e^{-(t-s)a(t)}
\leq C \eta(t)^{-2} \Vert w \Vert^2\end{aligned}$$ provided we take $2\alpha\leq {3\over 2}$ in the definition of $\eta(t)$.

Let now $q=q'$. Call the summands in and $v_{ar}$ and $v_{br}$. We consider first in the integral in $s\geq s_0$ with $ \eta(s_0)^{-1} = 4\eta(t)^{-1}$. Using $$\label{var}
|v_{ar}(s,q,q)|_t\leq Ce^{-2\beta_s|q-r|}\eta(s)^{-2}\Vert w \Vert^2$$ we get $$\label{vat}
\int_{s_0}^t |v_a(s,q,q) |_tds\leq C \eta(t)^{-2}\Vert w \Vert^2.$$ In the same way we get $$\label{vbt}
\int_{s_0}^t |v_{b}(s,q,q)|_t\leq C \Vert w \Vert^2 \int_{s_0}^t e^{s/2-2s}\eta(s)^{2}
ds \leq C \eta(t)^{-2} \Vert w \Vert^2$$

Finally, consider $v_a(s,q,q)$ for $s\leq s_0$. For such $s$ the $r=q$ term is zero. Indeed, ${\mathop{\hbox{{\rm Supp}}}}\dot\gamma_s \subset [\eta(s)^{-1}, 2\eta(s)^{-1}]$, but $|\omega  q + \kappa| \leq 4\eta(t)^{-1}\leq\eta(s)^{-1}$. For $r\neq q$ we have $$\label{dg}
|\partial^i_{\kappa} v_{ar}| \leq Ce^{-2\beta_s|q-r|}\eta(s)^{-2+i}\Vert w \Vert^2$$ for $i \leq 2$ and $|\kappa| \leq \eta(s)^{-1}$. We conclude, since $|q-r|>C\eta(s)^{1\over \nu}$ that $$\label{iva}
| \partial^i_\kappa \int_0^{s_0} v_a(s,q,q)ds | \leq C \Vert w \Vert^2$$ which holds for $|\kappa| \leq 4\eta(t)^{-1}$ since $ \eta(s_0)^{-1}\geq 4\eta(t)^{-1}$. In a similar manner we bound, for $|\kappa| \leq 4 \eta(t)^{-1}$, $$\label{ivb}
| \partial^i_\kappa \int_0^{s_0} v_b(s,q,q)ds | 
\leq C \Vert w \Vert^2$$ By the symmetry $$\label{symm}
v_1(s,q,q, \kappa) = v_1(s,0,0; \omega q + \kappa).$$ Let $\Psi(q,\kappa)=\int_0^{s_0}v_1(s,q,q,\kappa)$. Since $\partial_\kappa^i\Psi(0,0)=0$ for $i<2$ we have $$\label{Dint}
\Psi(q,\kappa)=\int_0^{\omega \cdot q + \kappa} 
\partial^2_{\kappa}\Psi (0, \kappa')(\omega q+\kappa-\kappa' )d\kappa'.$$ Since $\Psi(q,\kappa)$ satisfies for $|\kappa| \leq 4 \eta(t)^{-1}$ and $|\omega \cdot q + \kappa| \leq 3 \eta(t)^{-1}+ \eta(t)^{-1}$, is defined and by we infer that $$\label{Dbound}
|\Psi|_t \leq C \Vert w \Vert^2 \eta(t)^{-2}.$$ Thus , and give $$\label{phi1final}
\Vert \Phi_1\Vert_t \leq C \eta(t)^{-2} \Vert w \Vert^2.$$

To finish the proof we need to deal with $n >1$ in . Very crude bounds suffice. Using $$\label{exp}
e^{\beta_t(q-q' \sum p_i)} \leq e^{\beta_s(|q-r-q' - \sum p'_i | + |r - \sum p''_i|)}$$ for the first term and $|q-r-\sum p'_i | + |r -q'- \sum p''_i|$ in the exponent in the second term we get $$\begin{aligned}
\label{phin}
\sum_q|\Phi_n|_t e^{\beta_t |q-q'- \sum p_i|} & \leq \sum_{k=1}^{n+1} k \int_0^t \Vert w_k\Vert_s \Vert w_{n+1-k}\Vert_s \eta(s)^2 ds
 \leq C \Vert w \Vert^2\rho^{-n} \int_0^t   [ \eta(s)^{-2} n^{-2} e^{(n - 3/2)s}  \notag \\
 &+ (n+1)^{-1}\rho^{-1}  e^{(n - 1/2)s}e^{-2s} \quad+\rho^{-1} \sum_{k=2}^n \frac{1}{k} \frac{1}{(n+1-k)^2} e^{(n-2)s}] \eta(s)^2 ds \notag \\
& \leq C \Vert w \Vert^2 \rho^{-n-1}e^{(n-3/2)t} n^{-2}[ \frac{\rho}{n} + 
1 + n\sum_{k=2}^n \frac{1}{k} \frac{1}{(n+1-k)^2} ]\end{aligned}$$ provided $\eta(s)^2 e^{-s/2} < C$ which holds if $\alpha < \frac{1}{4}$. The term in the squared parenthesis is bounded in $n$ and thus $$\label{phinfinal}
\Vert \Phi_n \Vert_t \leq C n^{-2}\rho^{-n-1}e^{(n-3/2)t} \Vert w \Vert ^2.$$ Eqs. , , imply that $\Phi$ maps the ball $B_u( r)$ to $B_0( r)$ with $r=
 C\rho^{-1}\lambda^2$ if $\rho^{-1}\lambda$ is small enough. Contraction in this ball is similar.

The solution
============

Let us finally solve the $f$-equation , written as $$\label{Eqf}
f_n(t,q, {\boldsymbol{p}})  = a\delta_{n,1} \delta_{q,p} \mathbbm{1} + \sum_{k=1}^{n+1} \int_0^t \sum_{q'} k f_k(s,q,q',{\boldsymbol{p}}') \dot\gamma_s(\omega \cdot q') w_{n+1-k}(s,q',{\boldsymbol{p}}'') ds
 \equiv g_n +  {\mathcal{F}}_n$$ where $a=1$ for but we keep $a\geq 0$ for later purposes. Let $$\label{fnormt}
\Vert f_n \Vert_t = \sup_{{\boldsymbol{p}}\in \Lambda_t^n} \sum_{q \in {\mathbb{Z}}^d} e^{\beta_t |q- \sum p_i |}|f_n(q, {\boldsymbol{p}})|$$ and $$\label{fnorm}
\Vert f \Vert = \sup_{t \geq 0} \left( \Vert f_0 \Vert_t + \Vert f_1\Vert_t + \sup_{n \geq 2} n^{2} \rho^{n}e^{(1-n)t} \Vert f_n \Vert_t \right).$$ Note that differs from in the constraint on $q$ and the absence of $\kappa$. We use abusively the same notation. We have

\[final\] Equation has a unique solution in the ball $\Vert f - g \Vert \leq Ca \rho^{-1}\Vert w \Vert$.

We proceed as in the proof of Proposition \[mainprop\] to get $$\label{Fhigh}
\Vert {\mathcal{F}}_{n} \Vert_t \leq \sum_{k=1}^{n+1} \int_0^t k \Vert f_{k}(s) \Vert_s \Vert \dot\gamma_s \Vert \Vert w_{n+1-k} \Vert_s ds.$$ For $n=0$ this gives $$\label{Fzero}
\Vert {\mathcal{F}}_{0}\Vert \leq C \Vert w \Vert \Vert f \Vert \int_0^t \eta(s)^2 e^{-2s} ds \leq C \Vert w \Vert \Vert f \Vert$$ and for $n > 1$ $$\begin{aligned}
\Vert {\mathcal{F}}_{n} \Vert_t & \leq  C\rho^{-n-1} \Vert w \Vert \Vert f \Vert
\int_0^t  \Big[\rho n^{-2}e^{(n-3/2)s} + (n+1)^{-1}e^{ns}e^{-2s} 
 + \rho n^{-1}e^{(n-1)s}\eta(s)^{-2} \notag \\
& \quad+ \sum_{k=2}^{n-1} k^{-1} e^{(k-1)s} e^{(n+1-k-3/2)s}\Big]\eta(s)^2ds 
 \leq  C\Vert w \Vert \Vert f \Vert \rho^{-n-1}n^{-2}e^{(n-1)t}. \label{Fmorethanone}\end{aligned}$$ For $n=1$, the $k=2$ term in is bounded by $$\label{Fzero}
 C\rho^{-1} \Vert w \Vert \Vert f \Vert \int_0^t  e^{s}e^{-2s} \eta(s)^2ds \leq C\rho^{-1}  \Vert w \Vert \Vert f \Vert.$$ For the $k=1$ term we need to improve from a bit and extract as before the factor $e^{(\beta_t - \beta_s)|q-q'|} $, bounded by $ e^{-(t-s)\alpha(s)}$ to get the bound $$\label{Fone}
C \Vert w \Vert \Vert f \Vert \int_0^t  \eta(s)^{-2} \eta(s)^2e^{-(t-s)\alpha(s)}ds \leq C\Vert w \Vert \Vert f \Vert$$ for the $q=p\neq q'$ terms in the eq. . For $q=p=q'$ we observe that $\dot\gamma_s(\omega q)$ vanishes for $3\eta(s)\leq\eta(t)$ since $q\in \Lambda_t$. Thus the integral in is restriceted to $s>t-{\mathcal O}(1)$ and therefore bounded by again.

So, altogether we have $$\label{F1}
\Vert {\mathcal{F}}_{1} \Vert_t \leq C\rho^{-1}  \Vert w \Vert \Vert f \Vert.$$

We may now prove Theorem 1. First we note the limit exists and is real analytic. Indeed, let $x(t,q)=f_0(t,q)$. From Proposition 2 we deduce $$\label{conv}
\Vert f_0(t)-f_0(s)\Vert_t ={\mathcal O}(e^{-2s}\eta(s)^2)$$ i.e. $x(t,q)$ converge uniformly to an exponentially decaying sequence $x(q)$. We will shortly prove that $x(t)$ solves the equations $$\label{apr}
x(t)=\gamma_tu(x(t)), \ \ u(x(t))=w(t,0).$$ The first equation implies that $x(t,0)=0$ and for $|\omega q|\geq 2\eta(t)^{-1}$, $(\omega q)^2x(t,q)=u(q,x(t))$. Hence, for $q\neq 0$ $$\label{apr1}
(\omega q)^2x(q)=u(q,x)$$ and $x(0)=0$. The second equation combined with yields $u(0,x(t))=0$ and so by limits the second eq. follows. Our solution is clearly analytic in $\lambda$ and thus unique (up to the translations) since the equation determines uniquely the Taylor coefficients of its analytical solution.

We still need to prove . Consider the functionals $$\label{phi}
\phi(t,y)=y+\gamma_tu(f(t,y)),\ \ \ \psi(t,y)=u(f(t,y)).$$ $\phi$ and $\psi$ are analytic in $y$ in a neighbourhood of $0$ in $\ell^1(\Lambda_t)$. It suffices to show $\phi=f$ and $\psi=w$ since follows at $y=0$. Differentiating and using we get $$\label{phidot}
\dot\phi=\dot\gamma\psi+\gamma Du\dot f=\dot\gamma\psi+
\gamma DuDf\dot\gamma w=D\phi\dot\gamma w+\dot\gamma\rho$$ where $\rho=\psi-w$. Similar calculation gives $ \dot\psi=D\psi\dot\gamma w$ and combining with we get $$\label{rhodot}
\dot\rho=D\rho\dot\gamma w.$$ Initial conditions are $\phi(0,y)=f(0,y)$ and $\rho(0,y)=0$. Applying Proposition 2 with $a=0$ we get that $\rho=0$ and then with $a=1$ that $\phi=0$. Hence follows.

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