$$H \circ \Phi(\theta,I)=H_*(I)+R_*(\theta,I), \quad H_* \in G^{\beta}_{L_2}(B_{\bar{r}}), \quad R_* \in G^{\alpha,\beta}_{L_1,L_2}(\T^n \times B_{\bar{r}})$$ and with the following properties:

-   for any $(\theta,I) \in \T^n \times B_{\bar{r}}$ (resp. for any $(\theta',I') \in \Phi(\T^n \times B_{\bar{r}})$), we have $\Pi_I\Phi(\theta,I)=\mathrm{Id}+O\left(||I||^2\right)$ (resp. $\Pi_{I'}\Phi(\theta',I')=\mathrm{Id}+O\left(||I'||^2\right)$), where $\Pi_I$ (resp. $\Pi_{I'}$) denotes projection onto action space;

-   the Taylor series of $H_*$ at $I=0$ is given by the Birkhoff formal series $H_{\infty}$;

-   $\partial_I^l R_*(\theta,0)=0$ for all $\theta \in \T^n$ and all $l\in {\mathbb{N}}^n$: hence for any $r$ such that $0<2r < \bar{r}$, $$||R_*||_{\alpha,\beta,L_1,L_2,B_{2r}} \leq A\exp\left(-(2L_2r)^{-\frac{1}{\alpha(1+\tau)}}\right).$$

This statement is a direct consequence of Theorem 2 in [@MP09], to which we refer for a proof. The statement $(2)$ follows by construction of $H_*$, whereas $(3)$ follows from Stirling’s formula. Indeed, the assumption that $R_*$ is $(\alpha,\beta)$-Gevrey and flat at $I=0$, implies, by Taylor’s formula, that for any $m\in{\mathbb{N}}$: $$|\partial_\theta^k\partial_I^lR_*(\theta,I)|\leq ||R_*||_{\alpha,\beta,L_1,L_2,\bar{B}_{\bar{r}}}L_1^{|k|}L_2^{|l|}k!^\alpha l!^\beta m!^{\beta-1}(L_2||I||)^m.$$ Choosing $$m\sim (L_2||I||)^{-\frac{1}{\beta-1}}=(L_2||I||)^{-\frac{1}{\alpha(1+\tau)}},$$ by Stirling’s formula we eventually obtain $$|\partial_\theta^k\partial_I^lR_*(\theta,I)|\leq AL_1^{|k|}L_2^{|l|}k!^\alpha l!^\beta \exp\left(-(L_2\||I||)^{-\frac{1}{\alpha(1+\tau)}}\right)$$ where $A$ depends only on $||R_*||_{\alpha,\beta,L_1,L_2,\bar{B}_{\bar{r}}}$ and $n$. For $||I||<2r$, this proves $(3)$.

Les us now give an analogous statement for a Hamiltonian as in .

\[thmP\] Let $H_\varepsilon$ be as in , with $h \in G^\alpha_L(\bar{D})$ and $f \in G^\alpha_L(\T^n \times \bar{D})$, and 