V. I. Arnold’s “Global” KAM Theorem and Geometric Measure Estimates

Abstract

This paper continues the discussion started in [ 10 ] concerning Arnold’s legacy on classical KAM theory and (some of) its modern developments. We prove a detailed and explicit “global” Arnold’s KAM theorem, which yields, in particular, the Whitney conjugacy of a non-degenerate, real-analytic, nearly-integrable Hamiltonian system to an integrable system on a closed, nowhere dense, positive measure subset of the phase space. Detailed measure estimates on the Kolmogorov set are provided in case the phase space is: (A) a uniform neighbourhood of an arbitrary (bounded) set times the $d$ -torus and (B) a domain with $C^{2}$ boundary times the $d$ -torus. All constants are explicitly given.

APPENDIX A. PROOF OF THEOREM 1

In this appendix we provide the details needed to prove Arnold’s Global KAM Theorem (Theorem 1). The main point is the choice of the various parameters and sequences involved in the Newton-like procedure based on the iteration of a “KAM step” (in turn, based upon the original scheme by Arnold; compare [ 2 ] and its revisions in [ 12 ] and [ 10 ]). Although the main ideas are well known, some details are needed, especially in order to compute explicitly constants and to keep the optimal relation between ${\varepsilon}$ and ${\alpha}$ . Furthermore, the construction of the “integrating map” also requires a discussion. All this is done in the present appendix. By following ([ 12 ], Chap. 6), one gets the following:

General Step of the KAM Scheme

Lemma 1 (KAM step)

Let $r>0,0<2{\sigma}<s\leqslant 1$ , ${\mathscr{D}}_{\sharp}\subseteq{{\mathbb{R}}^{d}}$ be a non-empty, bounded domain. Consider the Hamiltonian parametrized by ${\varepsilon}\in{\mathbb{R}}$

$$H(y,x;{\varepsilon}):= K(y)+{\varepsilon}P(y,x),$$

where $K,P\in\mathcal{{\rm B}}_{r,s}({\mathscr{D}}_{\sharp})$ . Assume that Footnote

In the sequel, $K$ and $P$ stand for generic real-analytic Hamiltonians which later on will, respectively, play the roles of $K_{j}$ and $P_{j}$ , and $y_{0},r$ , the roles of $y_{j},r_{j}$ in the iterative step.

$$\displaystyle\det K_{yy}(y)\not=0,\qquad\qquad\qquad T(y):= K_{yy}(y)^{-1},\quad\forall y\in{\mathscr{D}}_{\sharp},$$

$$\displaystyle\|K_{yy}\|_{r,{\mathscr{D}}_{\sharp}}\leqslant\mathsf{M},\qquad\qquad\qquad\ \|T\|_{{\mathscr{D}}_{\sharp}}\leqslant\mathsf{L},$$

(A.1)

$$\displaystyle\|P\|_{r,s,{\mathscr{D}}_{\sharp}}\leqslant\mathsf{P},\qquad\qquad\qquad K_{y}({\mathscr{D}}_{\sharp})\subseteq{\Delta}^{\tau}_{\alpha}.$$

Fix ${\varepsilon}\neq 0$ and assume that

$${\lambda}\geqslant\log\left({\sigma}^{2{\nu}+d}\frac{{{\alpha}}^{2}}{{{\varepsilon}}{\mathsf{P}}\mathsf{M}}\right)\geqslant 1.$$

(A.2)

Let

$$\displaystyle\ell:= 4{\sigma}^{-1}{\lambda},\quad\ \ \check{r}\leqslant\frac{r}{32d\mathsf{L}\mathsf{M}},\quad\bar{r}\leqslant\min\left\{\frac{{\alpha}}{2d\mathsf{M}\ell^{\nu}},\check{r}\right\},$$

(A.3)

$$\displaystyle\tilde{r}:=\frac{\check{r}{\sigma}}{16d\mathsf{L}\mathsf{M}},\quad\quad\bar{s}:= s-\frac{2}{3}{\sigma},\quad s^{\prime}:= s-{\sigma},$$

and Footnote

Notice that ${{\mathfrak{p}}}\geqslant{\sigma}^{-d}\overline{{{\mathfrak{p}}}}\geqslant\overline{{{\mathfrak{p}}}}$ since ${\sigma}\leqslant 1$ . Notice also that $\mathsf{L}\mathsf{M}\geqslant 1$ , so that $\frac{16\mathsf{L}}{r\check{r}}{\sigma}^{-({\nu}+d)}>\frac{16\mathsf{L}}{r^{2}}\geqslant\frac{4}{\mathsf{M}r^{2}}$ .

$$\displaystyle{{\mathfrak{p}}}:=\mathsf{P}\max\left\{\frac{16\mathsf{L}}{r\bar{r}}{\sigma}^{-({\nu}+d)},\frac{\mathsf{C}_{4}}{{\alpha}\bar{r}}{\sigma}^{-2({\nu}+d)}\right\}.$$

Assume:

$${{\varepsilon}}{{{\mathfrak{p}}}}\leqslant\frac{\sigma}{3}.$$

(A.4)

Then there exists a diffeomorphism $G\colon{\mathbb{B}}_{\tilde{r}}({\mathscr{D}}_{\sharp}){\to}G({\mathbb{B}}_{\tilde{r}}({\mathscr{D}}_{\sharp}))$ , a symplectic change of coordinates

$$\phi^{\prime}={\mathtt{id}}+{\varepsilon}\tilde{\phi}:{\mathbb{B}}_{\bar{r}/2,s^{\prime}}({\mathscr{D}}_{\sharp}^{\prime})\to{\mathbb{B}}_{2r/3,\bar{s}}({\mathscr{D}}_{\sharp}),$$

(A.5)

such that

$$\left\{\begin{aligned} &\displaystyle H\circ\phi^{\prime}=: H^{\prime}=: K^{\prime}+{\varepsilon}^{2}P^{\prime}\ ,\\ &\displaystyle\partial_{y^{\prime}}K^{\prime}\circ G=\partial_{y}K,\quad\det\partial_{y^{\prime}}^{2}K^{\prime}\circ G\neq 0\quad\mbox{ on }{\mathscr{D}}_{\sharp},\end{aligned}\right.$$

(A.6)

with $K^{\prime}(y^{\prime}):= K(y^{\prime})+{\varepsilon}\widetilde{K}(y^{\prime}):= K(y^{\prime})+{\varepsilon}{\left\langle P(y^{\prime},\cdot)\right\rangle}$ . Moreover, letting $\left(\partial^{2}_{y^{\prime}}K^{\prime}(\mathsf{y}^{\prime})\right)^{-1}=: T(\mathsf{y}^{\prime})+{\varepsilon}\widetilde{T}(\mathsf{y}^{\prime})$ , $\mathsf{y}^{\prime}\in G({\mathscr{D}}_{\sharp})$ , the following estimates hold:

$$\left\{\begin{aligned} &\displaystyle\|\partial_{y^{\prime}}^{2}\widetilde{K}\|_{r/2,{\mathscr{D}}_{\sharp}}\leqslant\mathsf{M}{{\mathfrak{p}}},\qquad\|G-{\mathtt{id}}\|_{\tilde{r},{\mathscr{D}}_{\sharp}}\leqslant{\sigma}^{{\nu}+d}\bar{r}{{\varepsilon}}{{\mathfrak{p}}},\qquad\|\widetilde{T}\|_{{\mathscr{D}}_{\sharp}^{\prime}}\leqslant\mathsf{L}{{\mathfrak{p}}},\\ &\displaystyle\max\left\{\frac{\mathsf{C}_{12}}{\mathsf{C}_{4}}\|\overline{\mathsf{W}}\nabla\tilde{\phi}\overline{\mathsf{W}}^{-1}\|_{\bar{r}/2,s^{\prime},{\mathscr{D}}_{\sharp}^{\prime}},\|\mathsf{W}\tilde{\phi}\|_{\bar{r}/2,s^{\prime},{\mathscr{D}}_{\sharp}^{\prime}}\right\}\leqslant{\sigma}^{d}{{{\mathfrak{p}}}},\qquad\|P^{\prime}\|_{\bar{r}/2,s^{\prime},{\mathscr{D}}_{\sharp}^{\prime}}\leqslant{{\mathfrak{p}}}\mathsf{P},\end{aligned}\right.$$

(A.7)

where

$${\mathscr{D}}_{\sharp}^{\prime}:= G({\mathscr{D}}_{\sharp}),\quad\left(\partial^{2}_{y^{\prime}}K^{\prime}(\mathsf{y}^{\prime})\right)^{-1}=: T\circ G^{-1}(\mathsf{y}^{\prime})+{\varepsilon}\widetilde{T}(\mathsf{y}^{\prime}),\ \forall\mathsf{y}^{\prime}\in{\mathscr{D}}_{\sharp}^{\prime},$$

$$\mathsf{W}:=\mathop{\rm diag}\nolimits(\bar{r}^{-1}{\mathsf{1}}_{d},{\mathsf{1}}_{d}),\qquad\overline{\mathsf{W}}:=\mathop{\rm diag}\nolimits({\sigma}^{-{\tau}}\bar{r}^{-1}{\mathsf{1}}_{d},{\mathsf{1}}_{d}).$$

In the sequel, $K$ and $P$ stand for generic real-analytic Hamiltonians which later on will, respectively, play the roles of $K_{j}$ and $P_{j}$ , and $y_{0},r$ , the roles of $y_{j},r_{j}$ in the iterative step.

Notice that ${{\mathfrak{p}}}\geqslant{\sigma}^{-d}\overline{{{\mathfrak{p}}}}\geqslant\overline{{{\mathfrak{p}}}}$ since ${\sigma}\leqslant 1$ . Notice also that $\mathsf{L}\mathsf{M}\geqslant 1$ , so that $\frac{16\mathsf{L}}{r\check{r}}{\sigma}^{-({\nu}+d)}>\frac{16\mathsf{L}}{r^{2}}\geqslant\frac{4}{\mathsf{M}r^{2}}$ .

Implementation

As in [ 10 ], we shall separate the first step from the others. Let $H$ , $K$ , $P$ , ${\rho}$ , $s$ , $s_{*}$ , $\mathsf{W}$ , ${\mathsf{P}}$ , $\mathsf{M}$ , $\mathsf{L}$ , $\theta$ , $\epsilon$ be as in Section 2. Set

$$\displaystyle{\sigma}_{0}:=(s-s_{*})/2,\ \epsilon_{0}:=\epsilon,\ \theta_{0}:=\theta,\ r_{0}:={\rho},\ \mathsf{L}_{0}:=\mathsf{L},\ \mathsf{M}_{0}:=\mathsf{M},\ \mathsf{P}_{0}:=\mathsf{P},\ \mathsf{W}_{1}:=\mathsf{W},$$

$$\displaystyle{\lambda}_{0}:=\log\epsilon_{0}^{-1},\ {\lambda}_{*}:=\mathsf{C}_{7}{\sigma}_{0}^{-(4{\nu}+2d+1)}\theta_{0}^{2}{\lambda}_{0}^{2{\nu}},\ \theta_{*}:= 2^{2{\nu}+2d+1}\mathsf{C}_{5}^{2}\theta_{0}^{2},\ \ell_{0}:= 4{\sigma}_{0}^{-1}{\lambda}_{0},$$

$$\displaystyle K_{0}:= K,\ P_{0}:= P,\ H_{0}:= H,\quad{\mathscr{D}}_{0}:={{\mathscr{D}}^{*}}.$$

First step

Let

$$\displaystyle s_{1}:= s_{0}-{\sigma}_{0},\quad\check{r}_{1}:=\frac{r_{0}}{64d\theta_{0}},\quad\tilde{r}_{1}:=\frac{\check{r}_{1}{\sigma}_{0}}{32d\theta_{0}},\quad r_{1}:=\frac{1}{{2}}\min\left\{\frac{{\alpha}}{2d\sqrt{2}\mathsf{M}_{0}\ell_{0}^{{\nu}}},\check{r}_{1}\right\},$$

$$\displaystyle\mathsf{M}_{1}:=\left(1+\frac{{\sigma}_{0}}{3}\right)\mathsf{M}_{0},\quad\mathsf{L}_{1}:=\left(1+\frac{{\sigma}_{0}}{3}\right)\mathsf{L}_{0},\quad\hat{\epsilon}_{0}:=\mathsf{C}_{8}{\sigma}_{0}^{-(3{\nu}+2d+1)}\epsilon_{0}^{1/2},\quad\mathsf{P}_{1}:=\frac{\hat{\epsilon}_{0}\mathsf{P}_{0}}{{{\varepsilon}}},$$

$$\displaystyle{{\mathfrak{p}}}_{0}:=\mathsf{P}_{0}\max\left\{\frac{8\mathsf{L}_{0}}{r_{0}r_{1}}{\sigma}_{0}^{-({\nu}+d)},\frac{\mathsf{C}_{4}}{2{\alpha}r_{1}}{\sigma}_{0}^{-2({\nu}+d)}\right\}.$$

Lemma 2

Under the above assumptions and notations, if

$${\alpha}\leqslant\frac{\mathsf{C}_{4}}{16}\frac{r_{0}}{\mathsf{L}_{0}}\qquad\mbox{and}\qquad\max\left\{e\epsilon_{0},\hat{\epsilon}_{0}\right\}\leqslant 1,$$

(A.8)

then there exist ${\mathscr{D}}_{1}\subseteq{\mathscr{D}}$ , a real-analytic diffeomorphism

$$G_{1}\colon{\mathbb{B}}_{\tilde{r}_{1}}({{\mathscr{D}}^{*}}){\to}G_{1}({\mathbb{B}}_{\tilde{r}_{1}}({{\mathscr{D}}^{*}}))$$

and a real-analytic symplectomorphism

$$\phi_{1}:{\mathbb{B}}_{r_{1},s_{1}}({\mathscr{D}}_{1})\to{\mathbb{B}}_{r_{0},s_{0}}({\mathscr{D}}_{0})$$

(A.9)

such that

$$\displaystyle G_{1}({{\mathscr{D}}^{*}})={\mathscr{D}}_{1},$$

(A.10)

$$\displaystyle\partial_{y_{1}}K_{1}\circ G_{1}=\partial_{y}K_{0},$$

(A.11)

$$\displaystyle H_{1}:= H_{0}\circ\phi_{1}=: K_{1}+{\varepsilon}^{2}P_{1}\qquad\quad\ \quad\mbox{on }{\mathbb{B}}_{r_{1},s_{1}}({\mathscr{D}}_{1})$$

(A.12)

and Footnote

( A.17 ) follows trivially ( A.16 ) using Cauchy’s estimate.

$$\displaystyle{\mathscr{D}}_{1}\subseteq{\mathscr{D}}_{r_{1}},$$

(A.13)

$$\displaystyle\|\partial_{y_{1}}^{2}K_{1}\|_{r_{0}/4,{\mathscr{D}}_{1}}\leqslant\mathsf{M}_{1},\qquad\|T_{1}\|_{{\mathscr{D}}_{1}}\leqslant\mathsf{L}_{1},\qquad T_{1}:=(\partial_{y_{1}}^{2}K_{1})^{-1},$$

(A.14)

$$\displaystyle\|P_{1}\|_{r_{1},s_{1},{\mathscr{D}}_{1}}\leqslant\mathsf{P}_{1},$$

(A.15)

$$\displaystyle\|G_{1}-{\mathtt{id}}\|_{\tilde{r}_{1},{{\mathscr{D}}^{*}}}\leqslant 2{\sigma}_{0}^{{\nu}+d}r_{1}{{\varepsilon}}{{\mathfrak{p}}}_{0},$$

(A.16)

$$\displaystyle\|\partial_{z}G_{1}-{\mathsf{1}}_{d}\|_{\tilde{r}_{1}/2,{{\mathscr{D}}^{*}}}\leqslant 2^{5}d\mathsf{C}_{4}\sqrt{2}\theta_{0}{\sigma}_{0}^{{\tau}+d}\ell_{0}^{-{\nu}}{{\varepsilon}}{{\mathfrak{p}}}_{0},$$

(A.17)

$$\displaystyle\max\{\mathsf{C}_{12}\mathsf{C}_{4}^{-1}\|\overline{\mathsf{W}}_{1}\nabla(\phi_{1}-id)\overline{\mathsf{W}}_{1}^{-1}\|_{r_{1},s_{1},{\mathscr{D}}_{1}},\|\mathsf{W}_{1}(\phi_{1}-{\mathtt{id}})\|_{r_{1},s_{1},{\mathscr{D}}_{1}}\}\leqslant{\sigma}_{0}^{d}{{\varepsilon}}{{{\mathfrak{p}}}}_{0}.$$

(A.18)

( A.17 ) follows trivially ( A.16 ) using Cauchy’s estimate.

Second step, iteration and convergence

For a given $j\geqslant 1$ , define Footnote

Notice that $s_{j}\downarrow s_{*}$ and $r_{j}\downarrow 0$ .

$$\displaystyle{\sigma}_{j}:=\frac{{\sigma}_{0}}{2^{j}},\quad s_{j+1}:= s_{j}-{\sigma}_{j}=s_{*}+\frac{{\sigma}_{0}}{2^{j}},\quad\bar{s}_{j}:= s_{j}-\frac{2{\sigma}_{j}}{3},\quad\ell_{j}:= 4^{j}\ell_{0},$$

$$\displaystyle\mathsf{M}_{j+1}:=\mathsf{M}_{0}\prod_{k=0}^{j}(1+\frac{{\sigma}_{k}}{3})<\mathsf{M}_{0}\sqrt{2},\quad\mathsf{L}_{j+1}:=\mathsf{L}_{0}\prod_{k=0}^{j}(1+\frac{{\sigma}_{k}}{3})<\mathsf{L}_{0}\sqrt{2},$$

$$\displaystyle\epsilon_{j}:=\frac{\mathsf{M}_{0}{{\varepsilon}}^{2^{j}}\mathsf{P}_{j}}{{\alpha}^{2}},\quad\check{r}_{j+1}:=\frac{r_{j}}{64d\theta_{0}},\quad\tilde{r}_{j+1}:=\frac{\check{r}_{j+1}{\sigma}_{j}}{32d\theta_{0}},\quad r_{j+1}:=\frac{1}{{2}}\min\left\{\frac{{\alpha}}{2d\sqrt{2}\mathsf{M}_{0}\ell_{j}^{{\nu}}},\frac{r_{j}}{64d\theta_{0}}\right\},$$

$$\displaystyle\mathsf{P}_{j+1}:={\lambda}_{*}\theta_{*}^{j-1}\frac{\mathsf{M}_{0}\mathsf{P}_{j}^{2}}{{\alpha}^{2}},\quad\hat{\epsilon}_{j}:={\lambda}_{*}\theta_{*}^{j}\epsilon_{j},\quad\mathsf{W}_{j+1}:=\mathop{\rm diag}\nolimits\left((2r_{j+1})^{-1}{\mathsf{1}}_{d},{\mathsf{1}}_{d}\right),$$

$$\displaystyle\overline{\mathsf{W}}_{j+1}:=\mathop{\rm diag}\nolimits\left({\sigma}_{j}^{-{\tau}}(2r_{j+1})^{-1}{\mathsf{1}}_{d},{\mathsf{1}}_{d}\right),\quad{{\mathfrak{p}}}_{j}:=\mathsf{P}_{j}\max\left\{\frac{8\mathsf{L}_{0}\sqrt{2}}{r_{j}r_{j+1}}{\sigma}_{j}^{-({\nu}+d)},\frac{\mathsf{C}_{4}}{2{\alpha}r_{j+1}}{\sigma}_{j}^{-2({\nu}+d)}\right\}.$$

Observe that, for any $j\geqslant 1$ ,

$$\displaystyle\hat{\epsilon}_{j+1}={\lambda}_{*}\theta_{*}^{j+1}\epsilon_{j+1}={\lambda}_{*}\theta_{*}^{j+1}\frac{\mathsf{M}_{0}{{\varepsilon}}^{2^{j+1}}\mathsf{P}_{j+1}}{{\alpha}^{2}}={\lambda}_{*}\theta_{*}^{j+1}\frac{\mathsf{M}_{0}{{\varepsilon}}^{2^{j+1}}}{{\alpha}^{2}}{\lambda}_{*}\theta_{*}^{j-1}\frac{\mathsf{M}_{0}\mathsf{P}_{j}^{2}}{{\alpha}^{2}}\left({\lambda}_{*}\theta_{*}^{j}\epsilon_{j}\right)^{2}=\hat{\epsilon}_{j}^{2}$$

i.e.

$$\hat{\epsilon}_{j}=\hat{\epsilon}_{1}^{2^{j-1}}.$$

Notice that $s_{j}\downarrow s_{*}$ and $r_{j}\downarrow 0$ .

Lemma 3

Assume $(A.12)\div(A.15)$ with some ${\varepsilon}\neq 0$ and

$$\max\left\{e\epsilon_{0},2^{11}d^{2}\theta_{0}{\sigma}_{0}^{{\nu}+d}\hat{\epsilon}/3,2\mathsf{C}_{6}\theta_{0}\hat{\epsilon}_{1}\right\}\leqslant 1.$$

(A.19)

Then one can construct a sequence of real-analytic diffeomorphisms

$$G_{j}\colon{\mathbb{B}}_{\tilde{r}_{j}}({\mathscr{D}}_{j-1}){\to}G_{j}({\mathbb{B}}_{\tilde{r}_{j}}({\mathscr{D}}_{j-1})),\qquad j\geqslant 2,$$

and of real-analytic symplectic transformations

$$\phi_{j}:{\mathbb{B}}_{r_{j},s_{j}}({\mathscr{D}}_{j})\to{\mathbb{B}}_{r_{j-1},s_{j-1}}({\mathscr{D}}_{j-1}),$$

(A.20)

such that

$$\displaystyle G_{j}({\mathscr{D}}_{j-1})={\mathscr{D}}_{j}\subseteq{\mathscr{D}}_{r_{j}},$$

$$\displaystyle\partial_{y}K_{j+1}\circ G_{j+1}=\partial_{y}K_{j},$$

$$\displaystyle H_{j}:= H_{j-1}\circ\phi_{j}=: K_{j}+{\varepsilon}^{2^{j}}P_{j}\qquad\quad\ \mbox{on}\quad{\mathbb{B}}_{r_{j},s_{j}}({\mathscr{D}}_{j})$$

converge uniformly. More precisely, we have the following:

• (i) the sequence $G^{j}:= G_{j}\circ G_{j-1}\circ\cdots\circ G_{2}\circ G_{1}$ converges uniformly on ${{\mathscr{D}}^{*}}$ to a lipeomorphism $Y^{*}\colon{{\mathscr{D}}^{*}}\to{{\mathscr{D}}_{*}}:= Y^{*}({{\mathscr{D}}^{*}})\subseteq{\mathscr{D}}$ and $Y^{*}\in C^{\infty}_{W}({{\mathscr{D}}^{*}})$ .

• (ii) ${\varepsilon}^{2^{j}}\partial_{y}^{{\beta}}P_{j}$ converges uniformly on ${{\mathscr{D}}_{*}}\times\displaystyle{\mathbb{T}}^{d}_{s_{*}}$ to $0$ , for any ${\beta}\in{\mathbb{N}}_{0}^{d}$ ;

• (iii) $\phi^{j}:=\phi_{2}\circ\cdots\circ\phi_{j}$ converges uniformly on ${{\mathscr{D}}_{*}}\times{{\mathbb{T}}^{d}}$ to a symplectic transformation

  $$\phi^{*}\colon{{\mathscr{D}}_{*}}\times{{\mathbb{T}}^{d}}\overset{into}{\longrightarrow}{\rm B}_{r_{1}}({\mathscr{D}}_{1})\times{{\mathbb{T}}^{d}},$$

  with $\phi^{*}\in C^{\infty}_{W}({{\mathscr{D}}_{*}}\times{{\mathbb{T}}^{d}})$ and $\phi^{*}(y,\cdot)\colon{\mathbb{T}}^{d}_{s_{*}}\ni x\mapsto\phi^{*}(y,x)$ holomorphic, for any $y\in{{\mathscr{D}}_{*}}$ ;

• (iv) $K_{j}$ converges uniformly on ${{\mathscr{D}}_{*}}$ to a function $K_{*}\in C^{\infty}_{W}({{\mathscr{D}}_{*}})$ , with

  $$\displaystyle\partial_{y_{*}}K_{*}\circ Y^{*}=\partial_{y}K_{0}\quad\qquad\qquad\quad\quad\mbox{on}\quad{{\mathscr{D}}^{*}},$$
  $$\displaystyle\partial_{y_{*}}^{{\beta}}(H_{1}\circ\phi^{*})(y_{*},x)=\partial_{y_{*}}^{{\beta}}K_{*}(y_{*}),\quad\forall(y_{*},x)\in{{\mathscr{D}}_{*}}\times{{\mathbb{T}}^{d}},\forall{\beta}\in{\mathbb{N}}_{0}^{d}.$$

Finally, the following estimates hold for any $i\geqslant 2$ : ¹³

$$\displaystyle\|G_{i}-{\mathtt{id}}\|_{\tilde{r}_{i},{\mathscr{D}}_{i-1}}\leqslant 2r_{i}{\sigma}_{i-1}^{{\nu}+d}{{\varepsilon}}^{2^{i-1}}{{\mathfrak{p}}}_{i-1},$$

(A.21)

$$\displaystyle\|\partial_{z}G_{i}-{\mathsf{1}}_{d}\|_{\tilde{r}_{i}/2,{\mathscr{D}}_{i-1}}\leqslant 2^{5}d\theta_{0}{\sigma}_{i-1}^{{\tau}+d}{{\varepsilon}}^{2^{i-1}}{{\mathfrak{p}}}_{i-1},$$

(A.22)

$$\displaystyle\|P_{i}\|_{r_{i},s_{i},{\mathscr{D}}_{i}}\leqslant\mathsf{P}_{i}\ ,$$

(A.23)

$$\displaystyle\|\mathsf{W}_{2}(\phi^{i+2}-\phi^{i+1})\|_{r_{i+2},s_{i+2},\mathscr{D}_{i+2}}\leqslant\mathsf{a}_{2}\left(\mathsf{C}_{6}\theta_{0}^{\frac{1}{{4}}}\hat{\epsilon}_{1}\right)^{2^{i}},$$

(A.24)

$$\displaystyle|\mathsf{W}_{2}(\phi^{*}-{\mathtt{id}})|\leqslant\frac{2{\sigma}_{0}^{d+1}\hat{\epsilon}_{1}}{3\theta_{*}}\qquad\qquad\ \mbox{on}\quad{{\mathscr{D}}_{*}}\times{\mathbb{T}}^{d}_{s_{*}}\ ,$$

(A.25)

where

$$\mathsf{a}_{2}:=\mathbf{a}_{1}{\sigma}_{2}^{d}\|\mathsf{W}_{2}\phi_{2}\|_{r_{2},s_{2},\mathscr{D}_{2}}.$$

We can now complete the proof of Theorem 1 . First of all, observe that

$$(\log t)^{a}\leqslant\left(\frac{2a}{e}\right)^{a}\sqrt{t},\qquad\forall t\geqslant e,\quad\forall a>\frac{1}{{2}},$$

(A.26)

and from the proof we have

$$\displaystyle{{\varepsilon}}{{\mathfrak{p}}}_{0}(3\sigma_{0}^{-1})\overset{(A.8)}{\leqslant}6d\mathsf{C}_{4}\sqrt{2}{\sigma}_{0}^{-2({\nu}+d)-1}\frac{\mathsf{K}_{0}{{\varepsilon}}\mathsf{P}_{0}}{{\alpha}^{2}}\ell_{0}^{\nu}$$

(A.27)

$$\displaystyle\stackrel{{\scriptstyle(A.26)}}{{\leqslant}}\hat{\epsilon}_{0}\stackrel{{\scriptstyle(A.8)}}{{\leqslant}}1,$$

(A.28)

and, for $j\geqslant 1$ ,

$${{\varepsilon}}^{2^{j}}{{\mathfrak{p}}}_{j}(3\sigma_{j}^{-1})\leqslant{\hat{\epsilon}_{1}^{2^{j-1}}}/{\theta_{*}}.$$

(A.29)

Let $\phi_{*}:=\phi_{1}\circ\phi^{*}$ . Thus, uniformly on $\mathscr{D}_{*}\times{\mathbb{T}}^{d}_{s_{*}}$ , Footnote

Observe that ${\lambda}_{0}^{2{\nu}}\epsilon_{0}\stackrel{{\scriptstyle(A.19)}}{{\leqslant}}(4{\nu})^{2{\nu}}\sqrt{\epsilon_{0}}\stackrel{{\scriptstyle(A.19)}}{{\leqslant}}(4{\nu})^{2{\nu}}(2^{11}d^{2}\mathsf{C}_{5})^{-1/2}\theta_{0}^{-1}{\sigma}_{0}^{({\nu}+d+1)/2}$ .

$$\displaystyle|\mathsf{W}_{1}(\phi_{*}-{\mathtt{id}})|\leqslant|\mathsf{W}_{1}(\phi_{1}\circ\phi^{*}-\phi^{*})|+|\mathsf{W}_{1}(\phi^{*}-{\mathtt{id}})|$$

$$\displaystyle\leqslant\|\mathsf{W}_{1}(\phi_{1}-{\mathtt{id}})\|_{r_{1},s_{1},\mathscr{D}_{1}}+\|\mathsf{W}_{1}\mathsf{W}_{2}^{-1}\||\mathsf{W}_{2}(\phi^{*}-{\mathtt{id}})|$$

$$\displaystyle\leqslant{\sigma}_{0}^{d}{{\varepsilon}}{{{\mathfrak{p}}}}_{0}+\frac{2{\sigma}_{0}^{d+1}\hat{\epsilon}_{1}}{3\theta_{*}}$$

$$\displaystyle\stackrel{{\scriptstyle (A.26)+(A.27)}}{{\leqslant}}6d\mathsf{C}_{4}\sqrt{2}{\sigma}_{0}^{-(2{\nu}+2d+1)}\epsilon_{0}\ell_{0}^{\nu}+\left(\frac{{\nu}}{2e}\right)^{\nu}\mathsf{C}_{7}{\sigma}_{0}^{-(6{\nu}+4d+2)}\theta_{0}^{2}\ell_{0}^{{\nu}}\epsilon_{0}$$

$$\displaystyle\leqslant\mathsf{C}_{9}\theta_{0}^{2}\ell_{0}^{{\nu}}\epsilon_{0},$$

i. e., (2.10) . Moreover, setting $G_{0}:={\mathtt{id}}$ , we have, for any $i\geqslant 3$ ,

$$\displaystyle\|G^{i}-{\mathtt{id}}\|_{{{\mathscr{D}}^{*}}}\leqslant\sum_{j=0}^{i-1}\|G^{j+1}-G^{j}\|_{{{\mathscr{D}}^{*}}}=\sum_{j=0}^{i-1}\|G_{j}-{\mathtt{id}}\|_{{\mathscr{D}}_{j-1}}\overset{(A.21)+(A.16)}{\leqslant}2\sum_{j=0}^{i-1}r_{j+1}{\sigma}_{j}^{{\nu}+d}{{\varepsilon}}^{2^{j}}{{\mathfrak{p}}}_{j}$$

$$\displaystyle\leqslant 2r_{1}{\sigma}_{0}^{\nu}\sum_{j=0}^{\infty}{\sigma}_{j}^{d}{{\varepsilon}}^{2^{j}}{{\mathfrak{p}}}_{j}\stackrel{{\scriptstyle (A.28)+(A.29)}}{{\leqslant}}2r_{1}{\sigma}_{0}^{\nu}\cdot\mathsf{C}_{9}\theta_{0}^{2}\ell_{0}^{{\nu}}\epsilon_{0},$$

and then, passing to the limit, we get

$$\|Y^{*}-{\mathtt{id}}\|_{{{\mathscr{D}}^{*}}}\leqslant 2^{2{\tau}+1/2}d^{-1}\mathsf{C}_{9}{\sigma}_{0}^{\nu}\theta_{0}^{2}\frac{{{\varepsilon}}\mathsf{P}_{0}}{{\alpha}},$$

i. e., (2.8) . Now, observing that, for any $j\geqslant 1$ , $\nabla\phi^{j+1}=\nabla\phi^{j}\nabla\phi_{j+1}$ , $\|\overline{\mathsf{W}}_{j}\overline{\mathsf{W}}_{j+1}^{-1}\|=1$ and $\|\overline{\mathsf{W}}_{j+1}\overline{\mathsf{W}}_{j}^{-1}\|\leqslant\mathsf{C}_{5}\theta_{0}$ , we obtain

$$\displaystyle\|\overline{\mathsf{W}}_{1}(\nabla\phi^{j+1}-{\mathsf{1}}_{2d})\overline{\mathsf{W}}_{j+1}^{-1}\|_{*}\leqslant\left(\|\overline{\mathsf{W}}_{1}(\nabla\phi^{j}-{\mathsf{1}}_{2d})\overline{\mathsf{W}}_{j}^{-1}\|_{*}+1\right)\left(\|\overline{\mathsf{W}}_{j+1}(\nabla\phi_{j+1}-{\mathsf{1}}_{2d})\overline{\mathsf{W}}_{j+1}^{-1}\|_{*}+1\right)$$

$$\displaystyle\qquad-1$$

$$\displaystyle\leqslant\left(\|\overline{\mathsf{W}}_{1}(\nabla\phi^{j}-{\mathsf{1}}_{2d})\overline{\mathsf{W}}_{j}^{-1}\|_{*}+1\right)\left(\frac{\mathsf{C}_{4}}{\mathsf{C}_{12}}{\sigma}_{j}^{d}{{\varepsilon}}^{2^{j}}{{{\mathfrak{p}}}}_{j}+1\right)-1,$$

which iterated yields Footnote

Use: $e^{t}-1\leqslant te^{t}$ , for any $t\geqslant 0$ .

$$\displaystyle\|\overline{\mathsf{W}}_{1}(\nabla\phi^{j+1}-{\mathsf{1}}_{2d})\overline{\mathsf{W}}_{j+1}^{-1}\|_{*}\leqslant\prod_{j=1}^{\infty}\left(\frac{\mathsf{C}_{4}}{\mathsf{C}_{12}}{\sigma}_{j-1}^{d}{{\varepsilon}}^{2^{j-1}}{{{\mathfrak{p}}}}_{j-1}+1\right)-1$$

$$\displaystyle\leqslant\exp\left(\sum_{j=1}^{\infty}\frac{\mathsf{C}_{4}}{\mathsf{C}_{12}}{\sigma}_{j-1}^{d}{{\varepsilon}}^{2^{j-1}}{{{\mathfrak{p}}}}_{j-1}\right)-1$$

$$\displaystyle\stackrel{{\scriptstyle(A.28)+(A.29)}}{{\leqslant}}\exp(\mathsf{C}_{4}\mathsf{C}_{9}\mathsf{C}_{12}^{-1}\theta_{0}^{2}\ell_{0}^{{\nu}}\epsilon_{0})-1$$

$$\displaystyle\leqslant\exp((4d)^{-1})\mathsf{C}_{4}\mathsf{C}_{9}\mathsf{C}_{12}^{-1}\theta_{0}^{2}\ell_{0}^{{\nu}}\epsilon_{0}\stackrel{{\scriptstyle(2.4)+(A.26)}}{{\leqslant}}\frac{1}{{4(18d^{3}+70)\theta}},$$

and letting $j\to\infty$ , we obtain

$$\|\partial_{x}u_{*}\|_{*}\leqslant\exp((4d)^{-1})\mathsf{C}_{4}\mathsf{C}_{9}\mathsf{C}_{12}^{-1}\theta_{0}^{2}\ell_{0}^{{\nu}}\epsilon_{0}\leqslant\frac{1}{{4(18d^{3}+70)\theta}},$$

(A.30)

i. e., (2.11) .

Observe that ${\lambda}_{0}^{2{\nu}}\epsilon_{0}\stackrel{{\scriptstyle(A.19)}}{{\leqslant}}(4{\nu})^{2{\nu}}\sqrt{\epsilon_{0}}\stackrel{{\scriptstyle(A.19)}}{{\leqslant}}(4{\nu})^{2{\nu}}(2^{11}d^{2}\mathsf{C}_{5})^{-1/2}\theta_{0}^{-1}{\sigma}_{0}^{({\nu}+d+1)/2}$ .

Use: $e^{t}-1\leqslant te^{t}$ , for any $t\geqslant 0$ .

Next, we show that $\mathop{\rm Lip}\nolimits_{\mathscr{D}^{*}}(Y^{*}-{\mathtt{id}})<1$ , which will imply that Footnote

See Proposition II.2 in [ 20 ].

$Y^{*}\colon\mathscr{D}^{*}\overset{onto}{\longrightarrow}\mathscr{D}_{*}$ is a lipeomorphism. Observe first that, for any $j\geqslant 1$ , $0<r<\tilde{r}_{j}/2$ , $\mathsf{y}_{j-1}\in\mathscr{D}_{j-1}$ and any $y\in{\mathbb{B}}_{r}(\mathsf{y}_{j-1})$ , we have

$$|G_{j}(y)-G_{j}(\mathsf{y}_{j-1})|\leqslant|(G_{j}(y)-y)-(G_{j}(\mathsf{y}_{j-1})-\mathsf{y}_{j-1})|+|y-\mathsf{y}_{j-1}|\overset{(A.22)+(A.19)}{\leqslant}\frac{1}{{2}}|y-\mathsf{y}_{j-1}|+r<2r,$$

so that

$$G_{j}({\mathbb{B}}_{r}(\mathscr{D}_{j-1}))\subseteq{\mathbb{B}}_{2r}(G_{j}(\mathscr{D}_{j-1}))={\mathbb{B}}_{2r}(\mathscr{D}_{j}).$$

(A.31)

Thus, as the sequence $\tilde{r}_{j}$ is strictly decreasing, for any $j\geqslant k\geqslant 1$ , $G^{k}$ is well-defined on ${\mathbb{B}}_{2^{-j-1}\tilde{r}_{j+1}}(\mathscr{D}_{0})$ and we have

$$\displaystyle G^{k}({\mathbb{B}}_{2^{-j-1}\tilde{r}_{j+1}}(\mathscr{D}_{0}))\overset{(A.31)}{\subseteq}G_{j}\circ\cdots\circ G_{2}({\mathbb{B}}_{2^{-j}\tilde{r}_{j+1}}(\mathscr{D}_{1}))\overset{(A.31)}{\subseteq}\cdots\overset{(A.31)}{\subseteq}{\mathbb{B}}_{2^{k-j-1}\tilde{r}_{j+1}}(\mathscr{D}_{k})\subseteq$$

(A.32)

$$\displaystyle\subseteq{\mathbb{B}}_{2^{-1}\tilde{r}_{k+1}}(\mathscr{D}_{k}).$$

Therefore, for any $j\geqslant 2$ ,

$$\displaystyle\|G^{j}-{\mathtt{id}}\|_{L,{\mathbb{B}}_{2^{-j-1}\tilde{r}_{j+1}}(\mathscr{D}^{*})}+1=\|(G_{j}-{\mathtt{id}})\circ G^{j-1}+(G^{j-1}-{\mathtt{id}})\|_{L,{\mathbb{B}}_{2^{-j-1}\tilde{r}_{j+1}}(\mathscr{D}_{0})}+1$$

$$\displaystyle\leqslant(\|G_{j}-{\mathtt{id}}\|_{L,G^{j-1}({\mathbb{B}}_{\frac{\tilde{r}_{j+1}}{2^{j+1}}}(\mathscr{D}_{0}))}+1)(\|G^{j-1}-{\mathtt{id}}\|_{L,{\mathbb{B}}_{\frac{\tilde{r}_{j+1}}{2^{j+1}}}(\mathscr{D}_{0})}+1)$$

$$\displaystyle\stackrel{{\scriptstyle(A.32)}}{{\leqslant}}(\|G_{j}-{\mathtt{id}}\|_{L,{\mathbb{B}}_{\frac{\tilde{r}_{j}}{2}}(\mathscr{D}_{j-1})}+1)(\|G^{j-1}-{\mathtt{id}}\|_{L,{\mathbb{B}}_{\frac{\tilde{r}_{j}}{2^{j}}}(\mathscr{D}_{0})}+1)$$

$$\displaystyle=(\|\partial_{z}G_{j}-{\mathsf{1}}_{d}\|_{\tilde{r}_{j}/2,\mathscr{D}_{j-1}}+1)(\|G^{j-1}-{\mathtt{id}}\|_{L,{\mathbb{B}}_{\frac{\tilde{r}_{j}}{2^{j}}}(\mathscr{D}_{0})}+1)$$

$$\displaystyle\overset{(A.22)+(A.17)}{\leqslant}(2^{5}d\theta_{0}{\sigma}_{j-1}^{{\tau}+d}{{\varepsilon}}^{2^{j-1}}\mathsf{L}_{j-1}+1)(\|G^{j-1}-{\mathtt{id}}\|_{L,{\mathbb{B}}_{\frac{\tilde{r}_{j}}{2^{j}}}(\mathscr{D}^{*})}+1),$$

which iterated leads to Footnote

Use, again, $e^{t}-1\leqslant te^{t},\ \forall t\geqslant 0$ , and $2^{5}d\mathsf{C}_{4}\sqrt{2}\theta_{0}{\sigma}_{0}^{{\tau}+d}\ell_{0}^{-{\nu}}{{\varepsilon}}\mathsf{L}_{0}\stackrel{{\scriptstyle(A.27)}}{{\leqslant}}2^{7}d^{2}\mathsf{C}_{4}^{2}\theta_{0}{\sigma}_{0}^{-({\nu}+d+1)}\epsilon_{0}\stackrel{{\scriptstyle(A.19)}}{{<}}(32d)^{-1}$ .

$$\displaystyle\|G^{j}-{\mathsf{1}}_{d}\|_{L,\mathscr{D}^{*}}\leqslant-1+(1+(32d)^{-1})\prod_{i=2}^{\infty}(2^{5}d\theta_{0}{\sigma}_{j-1}^{{\tau}+d}{{\varepsilon}}^{2^{i-1}}\mathsf{L}_{i-1}+1)$$

$$\displaystyle\leqslant-1+\exp((32d)^{-1}+2^{5}d\theta_{0}\sum_{i=1}^{\infty}{\sigma}_{i}^{{\tau}+d}{{\varepsilon}}^{2^{i}}\mathsf{L}_{i})$$

$$\displaystyle\leqslant-1+\exp((32d)^{-1}+2^{5}d\theta_{0}{\sigma}_{1}^{{\tau}}\sum_{i=1}^{\infty}{\sigma}_{i}^{d}{{\varepsilon}}^{2^{i}}\mathsf{L}_{i})$$

$$\displaystyle\leqslant-1+\exp\left((32d)^{-1}+2^{5}d\theta_{0}{\sigma}_{1}^{{\tau}}\frac{2{\sigma}_{0}^{d+1}\hat{\epsilon}_{1}}{3\theta_{*}}\right)$$

$$\displaystyle\stackrel{{\scriptstyle(A.19)}}{{<}}\ -1+\exp\left((32d)^{-1}+(32d)^{-1}\right)\leqslant e^{1/(16d)}/(16d)<\frac{1}{{4d}}.$$

(A.33)

Hence, letting $j\to\infty$ , we find that $Y^{*}$ is Lipschitz continuous, with $\mathop{\rm Lip}\nolimits_{\mathscr{D}^{*}}(Y^{*}-{\mathtt{id}})$ satisfying (2.9) as

$$2^{5}d\mathsf{C}_{4}\sqrt{2}\theta_{0}{\sigma}_{0}^{{\tau}+d}\ell_{0}^{-{\nu}}{{\varepsilon}}\mathsf{L}_{0}+\sum_{j\geqslant 2}2^{5}d\theta_{0}{\sigma}_{j-1}^{{\tau}+d}{{\varepsilon}}^{2^{j-1}}\mathsf{L}_{j-1}\stackrel{{\scriptstyle(A.19)}}{{\leqslant}}{c_{2}}\theta^{3}(s-s_{*})^{-1}\frac{\mathsf{M}{{\varepsilon}}\mathsf{P}}{{\alpha}^{2}}\left(\log\frac{{\alpha}^{2}}{\mathsf{M}{{\varepsilon}}\mathsf{P}}\right)^{\nu}.$$

Next, we show that $\phi_{*}\in C^{\infty}_{W}(\mathscr{D}_{*}\times{{\mathbb{T}}^{d}})$ . For any $n,j\geqslant 1$ , we have

$$\displaystyle\|G^{n+j}-G^{j}\|_{\mathscr{D}^{*}}\leqslant\sum_{k=j}^{n+j-1}\|G^{k+1}-G^{k}\|_{\mathscr{D}^{*}}$$

$$\displaystyle\stackrel{{\scriptstyle(A.21)}}{{\leqslant}}\sum_{k=j}^{n+j-1}2r_{k+1}{\sigma}_{k}^{{\nu}+d}{{\varepsilon}}^{2^{k}}\mathsf{L}_{k}$$

$$\displaystyle\leqslant 2r_{j+1}{\sigma}_{j}^{{\nu}}\sum_{k\geqslant 1}{\sigma}_{k}^{d}{{\varepsilon}}^{2^{k}}\mathsf{L}_{k}$$

$$\displaystyle\leqslant 2r_{j+1}{\sigma}_{j}^{{\nu}}\frac{2{\sigma}_{0}^{d+1}\hat{\epsilon}_{1}}{3\theta_{*}}$$

$$\displaystyle\stackrel{{\scriptstyle(A.19)}}{{<}}{\sigma}_{j}^{{\nu}}\tilde{r}_{j+1}.$$

Now, letting $n\to\infty$ , we get

$$\|Y^{*}-G^{j}\|_{\mathscr{D}^{*}}<{\sigma}_{j}^{{\nu}}\tilde{r}_{j+1}<\frac{\tilde{r}_{j+1}}{4}.$$

(A.34)

Hence Footnote

Recall that, by definition, $G^{j}(\mathscr{D}^{*})=\mathscr{D}_{j}$ and $Y^{*}(\mathscr{D}^{*})=\mathscr{D}_{*}$ .

, for any $j\geqslant 1$ ,

$${\mathbb{B}}_{\frac{\tilde{r}_{j+1}}{4}}\big{(}G^{j}(\mathscr{D}^{*})\big{)}\stackrel{{\scriptstyle(A.34)}}{{\subseteq}}{\mathbb{B}}_{\frac{\tilde{r}_{j+1}}{2}}(\mathscr{D}_{*})\stackrel{{\scriptstyle(A.34)}}{{\subseteq}}{\mathbb{B}}_{\tilde{r}_{j+1}}(\mathscr{D}_{j})\subseteq{\mathbb{B}}_{r_{j}}(\mathscr{D}_{j}).$$

(A.35)

Therefore, for any $n\geqslant 1$ , we have

$$\displaystyle\sum_{j\geqslant 3}\|\mathsf{W}_{2}(\phi^{j}-\phi^{j-1})\|_{\tilde{r}_{j+1}/2,s_{j},\mathscr{D}_{*}}\left(\frac{\tilde{r}_{j+1}}{2}\right)^{-n}\stackrel{{\scriptstyle(A.35)}}{{\leqslant}}(2^{12}d^{2}\theta_{0}^{2})^{n}\sum_{j\geqslant 3}\|\mathsf{W}_{2}(\phi^{j}-\phi^{j-1})\|_{r_{j},s_{j},\mathscr{D}_{j}}(r_{j}{\sigma}_{j})^{-n}\overset{(A.24)}{\leqslant}(2^{12}d^{2}\theta_{0}^{2}{\sigma}_{1}r_{1})^{n}\mathsf{a}_{2}\sum_{j\geqslant 3}\left(\mathsf{C}_{6}\theta_{0}^{\frac{1}{4}}\hat{\epsilon}_{1}\right)^{2^{j-2}}\left(2\mathsf{a}_{1}\right)^{n(j-1)}<+\infty,$$

since, for $j$ sufficiently large,

$$\left(\mathsf{C}_{6}\theta_{0}^{\frac{1}{4}}\hat{\epsilon}_{1}\right)^{2^{j-1}}\left(2\mathsf{a}_{1}\right)^{nj}<\left(\sqrt{2}\mathsf{C}_{6}\theta_{0}^{\frac{1}{4}}\hat{\epsilon}_{1}\right)^{2^{j-1}}\stackrel{{\scriptstyle(A.19)}}{{\leqslant}}(1/\sqrt{2})^{2^{j-1}}.$$

Thus, letting $\Phi_{j}:=\phi_{1}\circ\phi^{j}$ and using the mean value theorem, we have

$$\displaystyle\sum_{j\geqslant 2}\|\mathsf{W}_{2}(\Phi_{j}-\Phi_{j-1})\|_{\tilde{r}_{j+1}/2,s_{j},\mathscr{D}_{*}}\left(\frac{\tilde{r}_{j+1}}{2}\right)^{-n}\leqslant\|\mathsf{W}_{2}\nabla\phi_{1}\mathsf{W}_{2}^{-1}\|_{{r}_{1},s_{1},\mathscr{D}_{1}}\times$$

$$\displaystyle\times\sum_{j\geqslant 3}\|\mathsf{W}_{2}(\phi^{j}-\phi^{j-1})\|_{\tilde{r}_{j+1}/2,s_{j},\mathscr{D}_{*}}\left(\frac{\tilde{r}_{j+1}}{2}\right)^{-n}$$

$$\displaystyle<\infty.$$

Consequently, writing

$$\Phi_{j}=(\Phi_{j}-\Phi_{j-1})+\cdots+(\Phi_{3}-\Phi_{2}),\qquad j\geqslant 2,$$

and invoking Lemma 8 (see Appendix B.7 ), we conclude that $\phi_{*}=\lim\Phi_{j}\in C^{\infty}_{W}(\mathscr{D}_{*}\times{{\mathbb{T}}^{d}})$ .Now we prove $Y^{*}\in C^{\infty}_{W}(\mathscr{D}^{*})$ analogously. For any $j\geqslant 2$ and $n\geqslant 1$ , we have

$$G^{j}=(G^{j}-G^{j-1})+\cdots+(G^{2}-G^{1}),$$

and, thanks to (A.31) , $G^{j+1}-G^{j}$ is well-defined on ${\mathbb{B}}_{2^{-j-2}\tilde{r}_{j+2}}(\mathscr{D}^{*})$ , for any $j\geqslant 1$ , so that

$$\displaystyle\sum_{j\geqslant 1}\|G^{j+1}-G^{j}\|_{\frac{\tilde{r}_{j+2}}{2^{j+2}},\mathscr{D}^{*}}\left(\frac{\tilde{r}_{j+2}}{2^{j+2}}\right)^{-n}=\sum_{j\geqslant 1}({2^{j+2}}{\tilde{r}_{j+2}}^{-1})^{n}\|(G_{j+1}-{\mathtt{id}})\circ G^{j}\|_{\frac{\tilde{r}_{j+2}}{2^{j+2}},\mathscr{D}^{*}}$$

$$\displaystyle\stackrel{{\scriptstyle(A.32)}}{{\leqslant}}\sum_{j\geqslant 2}({2^{j+2}}{\tilde{r}_{j+2}}^{-1})^{n}\|G_{j+1}-{\mathtt{id}}\|_{\tilde{r}_{j+1},\mathscr{D}_{j}}$$

$$\displaystyle\leqslant 2\sum_{j\geqslant 1}({2^{j+2}}{\tilde{r}_{j+2}}^{-1})^{n}r_{j+1}{\sigma}_{j}^{{\nu}+d}{{\varepsilon}}^{2^{j}}\mathsf{L}_{j}$$

$$\displaystyle\stackrel{{\scriptstyle(A.19)}}{{<}}\infty,$$

which proves that $Y^{*}\in C^{\infty}_{W}(\mathscr{D}^{*})$ .

See Proposition II.2 in [ 20 ].

Use, again, $e^{t}-1\leqslant te^{t},\ \forall t\geqslant 0$ , and $2^{5}d\mathsf{C}_{4}\sqrt{2}\theta_{0}{\sigma}_{0}^{{\tau}+d}\ell_{0}^{-{\nu}}{{\varepsilon}}\mathsf{L}_{0}\stackrel{{\scriptstyle(A.27)}}{{\leqslant}}2^{7}d^{2}\mathsf{C}_{4}^{2}\theta_{0}{\sigma}_{0}^{-({\nu}+d+1)}\epsilon_{0}\stackrel{{\scriptstyle(A.19)}}{{<}}(32d)^{-1}$ .

Recall that, by definition, $G^{j}(\mathscr{D}^{*})=\mathscr{D}_{j}$ and $Y^{*}(\mathscr{D}^{*})=\mathscr{D}_{*}$ .

Finally, we prove Kolmogorov’s non-degeneracy Footnote

See Appendix B.9 .

of the Kolmogorov tori $\phi_{*}(\mathscr{D}_{*}\times{\mathbb{T}}^{d})$ . Fix $y_{*}\in\mathscr{D}_{*}$ . Let $y_{0}:=(Y^{*})^{-1}(y_{*})$ and

$$\hat{\epsilon}:=\frac{1}{{4(18d^{3}+70)\theta}}.$$

Since $\|\partial_{x}u_{*}\|_{*}\stackrel{{\scriptstyle(2.11)}}{{\leqslant}}\hat{\epsilon}<1/2$ , the map $x\longmapsto x+u_{*}(y_{*},x)$ is a diffeomorphism of ${{\mathbb{T}}^{d}}$ . Letting

$$(\partial_{x}({\mathtt{id}}+u_{*})(y_{*},x))^{-1}=:{\mathsf{1}}_{d}+A(y_{*},x)\ ,$$

we have

$$\|A\|_{*}\leqslant 2\|\partial_{x}u_{*}\|_{*}\stackrel{{\scriptstyle(2.11)}}{{\leqslant}}2\hat{\epsilon}<1;\quad\|v_{*}\|_{*}\stackrel{{\scriptstyle(2.11)}}{{\leqslant}}\frac{\mathsf{C}_{9}\sqrt{2}}{4d}\theta^{2}\frac{{{\varepsilon}}\mathsf{P}}{{\alpha}}\stackrel{{\scriptstyle(2.4)}}{{\leqslant}}\frac{\mathsf{C}_{4}\mathsf{C}_{9}\sqrt{2}}{2^{5}d\mathsf{C}_{*}}{\rho}<\frac{{\rho}}{8}.$$

(A.36)

Moreover, write $K_{yy}(y_{*})=K_{yy}(y_{0})({\mathsf{1}}_{d}+K_{yy}(y_{0})^{-1}(K_{yy}(y_{*})-K_{yy}(y_{0})))$ and observe

$$\mathop{\rm dist}\nolimits(y_{0},\partial\mathscr{D})\geqslant{\rho}\quad\mbox{and}\quad|y_{*}-y_{0}|\stackrel{{\scriptstyle(2.8)+(2.4)}}{{\leqslant}}\frac{2^{{\tau}-5}\mathsf{C}_{4}\mathsf{C}_{9}\sqrt{2}}{d\mathsf{C}_{*}}{\rho}<\frac{{\rho}}{64d},$$

so that

$$\mathop{\rm dist}\nolimits(y_{*},\partial\mathscr{D})\geqslant\frac{{\rho}}{2}.$$

(A.37)

Thus, by the mean value theorem, we have

$$\displaystyle\|K_{yy}(y_{0})^{-1}(K_{yy}(y_{*})-K_{yy}(y_{0}))\|\stackrel{{\scriptstyle(A.37)}}{{\leqslant}}\mathsf{T}\frac{d^{2}\mathsf{K}}{{{\rho}/2}}|y_{*}-y_{0}|$$

$$\displaystyle{\stackrel{{\scriptstyle(2.8)}}{{\leqslant}}2^{{\tau}+11/2}d^{2}\mathsf{C}_{9}\theta^{3}\frac{\mathsf{K}{{\varepsilon}}\mathsf{P}_{0}}{{\alpha}{\rho}}\stackrel{{\scriptstyle(2.4)}}{{\leqslant}}\frac{2^{{\tau}+15/2}d^{2}\mathsf{C}_{4}\mathsf{C}_{9}}{\mathsf{C}_{*}}}\leqslant\frac{1}{{2}}.$$

Hence, $K_{yy}(y_{*})$ is invertible and $\|K_{yy}(y_{*})^{-1}\|\leqslant 2\|K_{yy}(y_{0})^{-1}\|\leqslant 2\mathsf{T}.$

See Appendix B.9 .

In [ 17 ] it is proven that the map

$$\phi^{y_{*}}(y,x):=(y_{*}+v_{*}(y_{*},x)+y+A^{T}y,x+u_{*}(y_{*},x))$$

is symplectic. Then

$$H\circ\phi^{y_{*}}(y,x)=E^{y_{*}}+{\omega}^{y_{*}}\cdot y+Q^{y_{*}}(y,x)$$

with

$$\displaystyle E^{y_{*}}=K(y_{*}),\quad{\omega}^{y_{*}}:= K_{y}(y_{0}),\quad\langle Q_{yy}^{y_{*}}(0,\cdot)\rangle=K_{yy}(y_{*})+{\left\langle\mathcal{M}\right\rangle},$$

$$\displaystyle\mathcal{M}:=\partial^{2}_{y}\bigg{(}K(y_{*}+v_{*}+y+A^{T}y)-\frac{1}{{2}}y^{T}K_{yy}(y_{*})y\bigg{)}\Big{|}_{y=0}+\partial^{2}_{y}({\varepsilon}P\circ\phi)\Big{|}_{y=0},$$

$$\displaystyle\|K_{yy}(y_{*})^{-1}\mathcal{M}\|_{*}\leqslant 2\mathsf{T}\mathcal{M}\stackrel{{\scriptstyle(A.36)}}{{\leqslant}}2(18d^{3}+70)\hat{\epsilon}\theta=1/2,$$

which shows that $\langle Q_{yy}^{y_{*}}(0,\cdot)\rangle$ is invertible.     $\square$

Remark 2

Here we list all the constants, which appear in the above proof and give an explicit expression for the constants $c_{k}$ ’s appearing in the statement of Theorem 1 . Recall that ${\tau}>d-1\geqslant 1$ and notice that all the $\mathsf{C}_{i}$ ’s are greater than $1$ and depend only upon $d$ and ${\tau}$ :

$$\displaystyle{\nu}:={\tau}+1,\quad\mathsf{C}_{0}:= 4\left(\frac{3}{2}\right)^{2{\nu}+d}\int_{{{\mathbb{R}}^{d}}}\left(|y|_{1}^{{\nu}}+d|y|_{1}^{2{\nu}}\right)e^{-|y|_{1}}dy,\quad\mathsf{C}_{1}:= 2\left(\frac{3}{2}\right)^{{\nu}+d}\int_{{{\mathbb{R}}^{d}}}|y|_{1}^{{\nu}}e^{-|y|_{1}}dy,$$

$$\displaystyle\mathsf{C}_{2}:= 2^{3d}d,\quad\mathsf{C}_{3}:= d^{2}\mathsf{C}_{1}^{2}+6d\mathsf{C}_{1}+\mathsf{C}_{2},\quad\mathsf{C}_{4}:=\max\left\{(1+d^{2})\mathsf{C}_{0},\mathsf{C}_{3}\right\},\quad\mathsf{C}_{5}:=\max\left\{2^{2{\nu}},2^{7}d\right\},$$

$$\displaystyle\mathsf{C}_{6}:=\left(2^{-d}\mathsf{C}_{5}\right)^{\frac{1}{{4}}},\quad\mathsf{C}_{7}:= 3\cdot 2^{4{\nu}+2d+3}d\sqrt{2}\max\left\{2^{2{\nu}+6}d,\mathsf{C}_{4}/2\right\}\mathsf{C}_{5},\quad\mathsf{C}_{8}:= 3\cdot 2^{3{\nu}+1}{\nu}^{\nu}e^{-{\nu}}d\mathsf{C}_{4}\sqrt{2},$$

$$\displaystyle\mathsf{C}_{9}:= 3\cdot 2^{-(4{\nu}+2d)}d\mathsf{C}_{4}\sqrt{2}+2^{-{\nu}}{\nu}^{\nu}e^{-{\nu}}\mathsf{C}_{7}\mathsf{C}_{8},\quad\mathsf{C}_{10}:= 2\left(\frac{3}{2}\right)^{{\nu}+d+1}\int_{{{\mathbb{R}}^{d}}}|y|_{1}^{{\nu}+1}e^{-|y|_{1}}dy,$$

$$\displaystyle\mathsf{C}_{11}:= 8\left(\frac{3}{2}\right)^{3{\tau}+d+2}\int_{{{\mathbb{R}}^{d}}}(2|y|_{1}^{{\tau}}+3|y|_{1}^{2{\tau}+1}+|y|_{1}^{3{\tau}+2})e^{-|y|_{1}}dy,\quad\mathsf{C}_{12}:=\max\{2\mathsf{C}_{10},2\mathsf{C}_{11},12\mathsf{C}_{0}\},$$

$$\displaystyle\mathsf{C}_{*}:=\max\{2^{11{\nu}+6d+4}{\nu}^{\nu}e^{-{\nu}}\mathsf{C}_{5}^{2}\mathsf{C}_{6}\mathsf{C}_{7}\mathsf{C}_{8},(2^{{\nu}/2-2d+2}(18d^{3}+70){\nu}^{\nu}e^{-{\nu}}\mathsf{C}_{4}\mathsf{C}_{9}\mathsf{C}_{12}^{-1})^{2},2^{{\tau}+8}d^{2}\mathsf{C}_{4}\mathsf{C}_{9}\sqrt{2}\}.$$

Then

$$\displaystyle c_{*}:=\mathsf{C}_{*},\phantom{AAAAAA.}c_{0}:= 2^{-4}\mathsf{C}_{4},\quad\quad\phantom{.}c_{1}:= 2^{{\tau}-1/2}d^{-1}\mathsf{C}_{9},$$

$$\displaystyle c_{2}:= 2^{2{\nu}+6}d\mathsf{C}_{9},\qquad c_{3}:=\mathsf{C}_{9},\phantom{AAAAA.}c_{4}:= e^{(4d)^{-1}}\mathsf{C}_{4}\mathsf{C}_{9}\mathsf{C}_{12}^{-1}.$$

(A.38)

Remark 3

There is a small flaw in [ 10 ]: The parameter Footnote

In the present remark, we will adopt the notations of [ 10 ].

$\mathsf{L}$ chosen in ([ 10 ], Lemma 1) is not big enough to ensure that the new perturbation $P^{\prime}$ and the symplectic change of coordinates $\phi$ are well-defined on $D_{\bar{r}/2,s^{\prime}}(\mathscr{D}_{\sharp}^{\prime})$ . The right choice is the following:

$$\displaystyle\mathsf{L}:=\mathsf{P}\max\Big{\{}\frac{40d\mathsf{T}^{2}\mathsf{K}}{r\bar{r}{\sigma}^{{\nu}+d}},\frac{2\mathsf{C}_{4}}{{\alpha}\bar{r}{\sigma}^{2({\nu}+d)}}\Big{\}},\quad\mathsf{W}:=\mathop{\rm diag}\nolimits(\bar{r}^{-1}{\mathsf{1}}_{d},{\mathsf{1}}_{d}),\quad\hat{\epsilon}_{0}:=\mathsf{C}_{9}{\sigma}_{0}^{-2({\nu}+d)-1}\epsilon_{0}\theta_{0}^{2}{\lambda}_{0}^{\nu},$$

$$\displaystyle\mathsf{L}_{j}:=\frac{\mathsf{P}_{j}}{r_{j+1}}\max\Big{\{}\frac{80d\sqrt{2}\mathsf{T}_{0}{\theta}_{0}}{r_{j}{\sigma}_{j}^{{\nu}+d}},\frac{\mathsf{C}_{4}}{{\alpha}{\sigma}_{j}^{2({\nu}+d)}}\Big{\}},\quad\mathsf{W}_{j}:=\mathop{\rm diag}\nolimits(2r_{j+1}^{-1}{\mathsf{1}}_{d},{\mathsf{1}}_{d}),\quad\hat{\epsilon}_{j+1}:=\frac{\mathsf{K}_{0}{\varepsilon}^{2^{j+1}}\mathsf{P}_{j+1}}{{\alpha}^{2}},$$

$$\displaystyle\mathsf{P}_{j+2}:={\lambda}_{*}{\theta}_{*}^{j+1}\frac{\mathsf{K}_{0}\mathsf{P}_{j+1}^{2}}{{\alpha}^{2}},\quad\hat{\epsilon}_{j+1}:={\lambda}_{*}{\theta}_{*}^{j+2}\epsilon_{j+1}.$$

Of course, one needs then to change accordingly (and in a straightforward way) the constants involved, as follows:

$$\displaystyle{\nu}:={\tau}+1$$

$$\displaystyle\mathsf{C}_{0}:= 4\sqrt{2}\left(\frac{3}{2}\right)^{2{\nu}+d}\int_{{{\mathbb{R}}^{d}}}\left(|y|_{1}^{{\nu}}+|y|_{1}^{2{\nu}}\right)e^{-|y|_{1}}dy,\quad\mathsf{C}_{1}:= 2\left(\frac{3}{2}\right)^{{\nu}+d}\int_{{{\mathbb{R}}^{d}}}|y|_{1}^{{\nu}}e^{-|y|_{1}}dy,$$

$$\displaystyle\mathsf{C}_{2}:= 2^{3d}d,\quad\mathsf{C}_{3}:=\left(d^{2}\mathsf{C}_{1}^{2}+6d\mathsf{C}_{1}+\mathsf{C}_{2}\right)\sqrt{2},\quad\mathsf{C}_{4}:=\max\left\{6d^{2}\mathsf{C}_{0},\mathsf{C}_{3}\right\},$$

$$\displaystyle\mathsf{C}_{5}:=\frac{3\cdot 2^{5}d}{5},\quad\mathsf{C}_{6}:={\max}\left\{2^{2{\nu}},\mathsf{C}_{5}\right\},\quad\mathsf{C}_{7}:= 3d\cdot 2^{4{\nu}+2}\sqrt{2}\max\left\{640d^{2},\mathsf{C}_{4}\right\},$$

$$\displaystyle\mathsf{C}_{8}:=\left(2^{-d}\mathsf{C}_{6}\right)^{1/8},\quad\mathsf{C}_{9}:= 3d\cdot 2^{2{\nu}+2}\sqrt{2}\max\left\{80d\sqrt{2},\mathsf{C}_{4}\right\},$$

$$\displaystyle\mathsf{C}_{10}:=(4{\nu}e^{-1})^{2{\nu}}\left(1+2^{4{\nu}+2d+2}({\nu}e^{-1})^{2{\nu}}\mathsf{C}_{6}^{2}\mathsf{C}_{7}\right)\mathsf{C}_{9}/(3d^{2}),\quad\mathsf{C}_{11}:=({5d\cdot 2^{3({\nu}+1)}})^{-1}{\mathsf{C}_{10}},$$

$$\displaystyle\mathsf{C}_{12}:= 2^{2(5{\nu}+4d+2)}\mathsf{C}_{6}^{2}\mathsf{C}_{7}\mathsf{C}_{8}\mathsf{C}_{9},\quad\mathsf{C}_{13}:=\mathsf{C}_{10}+\mathsf{C}_{11},\quad\mathsf{C}_{14}:=\mathsf{C}_{12},$$

$$\displaystyle\mathsf{C}_{15}:= 18d^{3}+70,\qquad\mathsf{C}_{16}:=(6{\nu}e^{-1})^{4{\nu}},\quad\mathsf{C}\ :=\max\{3\mathsf{C}_{10},\mathsf{C}_{13}\},$$

$$\displaystyle\mathsf{C}_{*}:=\max\left\{\mathsf{C}_{16}\mathsf{C}_{14}^{2/3},6\mathsf{C}_{15}\mathsf{C}_{16}\mathsf{C}^{2},2^{2(4{\nu}+2d+1)}\mathsf{C}_{16}\mathsf{C}_{9}^{2},\mathsf{C}_{10}^{2}\right\}.$$

The smallness condition $(14)$ and the estimate $(16)$ become, respectively,

$${\alpha}\leqslant\frac{r}{\mathsf{T}}\qquad\mbox{and}\qquad\epsilon\leqslant\epsilon_{*}:=\frac{(s-s_{*})^{a}}{\mathsf{C}_{*}\theta^{6}},$$

and

$$\max\Big{\{}\|u_{*}\|_{s_{*}},\ \|\partial_{x}u_{*}\|_{s_{*}},\frac{\mathsf{K}}{{\alpha}}(\log\epsilon^{-1})^{\nu}\|v_{*}\|_{s_{*}}\Big{\}}\leqslant\frac{\mathsf{C}\ \theta^{3}}{(s-s_{*})^{a/2}}\ \epsilon(\log\epsilon^{-1})^{\nu}\leqslant\frac{1}{4e},$$

where $a:= 6{\nu}+3d+2$ .

In the present remark, we will adopt the notations of [ 10 ].

APPENDIX B. TOOLS

B.1 Classical Estimates (Cauchy, Fourier)

Lemma 4 ( [ 7 ] )

Let $p\in{\mathbb{N}},r,s>0,y_{0}\in{{\mathbb{C}}^{d}}$ and let $f$ be a real-analytic function ${\mathbb{B}}_{r,s}(y_{0})$ with $\|f\|_{r,s}:=\sup_{{\mathbb{B}}_{r,s}(y_{0})}|f|<\infty.$ Then

(i) For any multi–index $(l,k)\in{\mathbb{N}}^{d}\times{\mathbb{N}}^{d}$ with $|l|_{1}+|k|_{1}\leqslant p$ and for any $0<r^{\prime}<r,0<s^{\prime}<s$ , Footnote

As usual, $\partial_{y}^{l}:=\frac{\partial^{|l|_{1}}}{\partial y_{1}^{l_{1}}\cdots\partial y_{d}^{l_{d}}},\forall y\in{{\mathbb{R}}^{d}},l\in{{\mathbb{Z}}^{d}}$ .

$$\|\partial_{y}^{l}\partial_{x}^{k}f\|_{r^{\prime},s^{\prime}}\leqslant p!\|f\|_{r,s}(r-r^{\prime})^{|l|_{1}}(s-s^{\prime})^{|k|_{1}}.$$

(ii) For any $k\in{{\mathbb{Z}}^{d}}$ and any $y\in{\mathbb{B}}_{r}(y_{0})$

$$|f_{k}(y)|\leqslant e^{-|k|_{1}s}\|f\|_{r,s}.$$

As usual, $\partial_{y}^{l}:=\frac{\partial^{|l|_{1}}}{\partial y_{1}^{l_{1}}\cdots\partial y_{d}^{l_{d}}},\forall y\in{{\mathbb{R}}^{d}},l\in{{\mathbb{Z}}^{d}}$ .

B.2 An Inverse Function Theorem

Theorem B.1

Let $D$ be a convex subset of ${{\mathbb{C}}^{d}}$ , $y_{0}\in D$ and let $f\in C^{1}(D,{{\mathbb{C}}^{d}})$ such that Footnote

$f^{\prime}$ being the Jacobian matrix of $f$ .

$\det f^{\prime}(y_{0})\not=0$ .Assume

$$\varrho:=\sup_{y\in D}\|{\mathsf{1}}-Tf^{\prime}(y)\|<1,\qquad\qquad T:=(f^{\prime}(y_{0}))^{-1}.$$

(B.1)

Then $\det f^{\prime}(y)\not=0$ , for each $y\in D$ and

$$\|(f^{\prime}(y))^{-1}\|\leqslant{\lambda}:=\frac{\|T\|}{1-\varrho}.$$

(B.2)

Moreover, $f$ is injective on $D$ and its inverse function $g:f(D)\stackrel{{\scriptstyle\rm onto}}{{\to}}D$ satisfies

$$\mathop{\rm Lip}\nolimits_{f(D)}(g)\leqslant{\lambda}.$$

(B.3)

Furthermore, if $D:={B_{r}(y_{0})}$ , ${\rho}:= r/{\lambda}$ and $z_{0}:= f(y_{0})$ , then

$${B_{\rho}(z_{0})}\subseteq f(D).$$

(B.4)

$f^{\prime}$ being the Jacobian matrix of $f$ .

Proof

For every $y\in D$ , we have $f^{\prime}(y)=f^{\prime}(y_{0})({\mathsf{1}}-A),$ where $A:={\mathsf{1}}-Tf^{\prime}(y)$ with $\|A\|\leqslant\varrho<1$ . Thus, $f^{\prime}(y)$ is invertible and

$$\|(f^{\prime}(y))^{-1}\|=\|(\sum_{n\geqslant 0}A^{n})T\|\leqslant\frac{\|T\|}{1-\varrho},$$

proving (B.2) . Now, consider the auxiliary map $F\colon D\ni y\longmapsto y-Tf(y)$ . We have $F\in C^{1}(D,{{\mathbb{C}}^{d}})$ and $\sup_{D}\|F^{\prime}\|\stackrel{{\scriptstyle(B.1)}}{{\leqslant}}\varrho.$ Thus, for every $y,{\bar{y}}\in D$ with $y\neq{\bar{y}}$ , we have

$$\displaystyle\|T\|\|f(y)-f({\bar{y}})\|\stackrel{{\scriptstyle(B.1)}}{{\geqslant}}\big{\|}T\big{(}f(y)-f({\bar{y}})\big{)}\big{\|}=\|(y-{\bar{y}})+(F({\bar{y}})-F(y))\|\geqslant\|y-{\bar{y}}\|-\|y-{\bar{y}}\|\sup_{D}\|F^{\prime}\|\geqslant\|y-{\bar{y}}\|(1-\varrho)\stackrel{{\scriptstyle(B.1)}}{{>}}0,$$

(B.5)

which shows that $f$ is injective on $D$ and, hence, that (B.3) holds.

To show (B.4) in the case $D:={B_{r}(y_{0})}$ and ${\rho}:= r/{\lambda}$ , fix $\eta\in{{\mathbb{C}}^{d}}$ with $\|\eta-z_{0}\|<{\rho}$ . We have to show that there exists ${\bar{y}}\in D$ such that $f({\bar{y}})=\eta$ . Define the map

$$\Phi:y\in D\mapsto\Phi(y):= y-T\big{(}f(y)-\eta\big{)}\in Y.$$

(B.6)

Then $\Phi$ is a contraction on $D$ . Indeed, $\Phi$ is $C^{1}$ , $\Phi^{\prime}(y)={\mathsf{1}}-Tf^{\prime}(y)$ and

$$\displaystyle\mathop{\rm Lip}\nolimits_{D}\Phi=\sup_{D}\|\Phi^{\prime}\|=\varrho<1.$$

(B.7)

Furthermore, $\Phi:D\to D$ , since, if $y\in D$ , then

$$\displaystyle\|\Phi(y)-y_{0}\|\leqslant\|\Phi(y)-\Phi(y_{0})\|+\|\Phi(y_{0})-y_{0}\|\stackrel{{\scriptstyle(B.7)}}{{\leqslant}}\varrho r+\|T\|\|\eta-z_{0}\|<\varrho r+\|T\|{\rho}=r.$$

Hence, by the contraction lemma, $\Phi$ has a (unique) fixed point ${\bar{y}}\in D$ , but $\Phi({\bar{y}})={\bar{y}}$ means $f({\bar{y}})=\eta$ .     $\square$

B.3 Internal Coverings

Given any non-empty subset $D$ of ${{\mathbb{R}}^{d}}$ , and given $r>0$ , an $r$ -internal covering of $D$ is a subset $P$ of $D$ such that $D\subseteq\bigcup_{y\in P}{\rm B}_{r}(y)$ ; the $r$ -internal covering number of $D$ , denoted $N_{r}^{\rm int}(D)$ , is the minimal cardinality of any $r$ -internal cover.

In [ 5 ] the following simple upper bound (having fixed the sup-norm in ${{\mathbb{R}}^{d}}$ ) on $N_{r}^{\rm int}(D)$ for bounded sets $D$ is given:

Lemma 5

Let $D\subseteq{\mathbb{R}}^{d}$ be a non-empty bounded set. Then, for any $r>0$ , one has Footnote

$[x]$ denotes the integer-part (or “floor”) function $\max\{n\in{\mathbb{Z}}|\ n\leqslant x\}$ , while $\lceil x\rceil$ denote the “ceiling function” $\min\{n\in{\mathbb{Z}}|\ n\geqslant x\}$ ; observe that $\lceil x\rceil\leqslant[x]+1$ .

$$N_{r}^{\rm int}(D)\leqslant\Big{(}\Big{[}\frac{\mathop{\rm diam}\nolimits D}{r}\Big{]}+1\Big{)}^{d}.$$

(B.8)

$[x]$ denotes the integer-part (or “floor”) function $\max\{n\in{\mathbb{Z}}|\ n\leqslant x\}$ , while $\lceil x\rceil$ denote the “ceiling function” $\min\{n\in{\mathbb{Z}}|\ n\geqslant x\}$ ; observe that $\lceil x\rceil\leqslant[x]+1$ .

For convenience of the reader, we reproduce here the elementary proof of the lemma.

Proof

It is enough to produce an $r$ -internal cover of $D$ with cardinality $N$ bounded by the right-hand side of (B.8) . If $D$ is a singleton, the claim is obvious with $N=1$ . Assume now that ${\delta}:=\mathop{\rm diam}\nolimits D>0$ , and let $M:=[{\delta}/r]+1$ and $z_{i}=\inf\{x_{i}|\ x\in D\}$ . Then $D\subseteq K:=z+[0,\delta]^{d}$ and one can find $0<r^{\prime}<r$ close enough to $r$ so that $\lceil{\delta}/r^{\prime}\rceil\leqslant[{\delta}/r]+1=M$ . Then one can cover $K$ with $M^{d}$ closed, contiguous cubes $K_{j}$ , $1\leqslant j\leqslant M^{d}$ , with edge of length $r^{\prime}$ . Let $j_{i}$ be the indices such that $K_{j_{i}}\cap D\neq\emptyset$ and pick a $y_{i}\in K_{j_{i}}\cap E$ ; let $1\leqslant N\leqslant M^{d}$ be the number of such cubes. Observe that, since we have chosen the sup-norm in ${\mathbb{R}}^{d}$ , we have $K_{j_{i}}\subseteq{\rm B}_{r}(y_{i})$ and (B.8) follows.     $\square$

B.4 Extensions of Lipschitz Continuous Functions

Here we recall a theorem due to Minty according to which a Lipschitz continuous function can be extended keeping unchanged both the sup-norm and the Lipschitz constant.

Theorem B.2 (G. J. Minty [ 13 ] )

Let $(V,{\left\langle\cdot,\cdot\right\rangle})$ be a separable inner product space, $\emptyset\neq A\subseteq V$ , $L>0,0<{\alpha}\leqslant 1$ and $g\colon A\to{{\mathbb{R}}^{d}}$ a $(L,{\alpha})$ -Lipschitz – Hölder continuous function on $A$ , namely, let $g$ satisfy

$$|g(x_{1})-g(x_{2})|_{2}\leqslant L\|x_{1}-x_{2}\|^{\alpha},\qquad\forall x_{1},x_{2}\in A,$$

(B.9)

where $\|\cdot\|$ denotes the norm on $V$ induced by the inner product. Then there exists a global $(L,{\alpha})$ -Lipschitz – Hölder continuous function Footnote

I. e., satisfying ( B.9 ) on $V$ .

$G\colon V\to{{\mathbb{R}}^{d}}$ such that $G|_{A}=g$ . Futhermore, $G$ can be chosen in such a way that $G(V)$ is contained in the closed convex hull of $g(A)$ . Hence, in particular,

$$\displaystyle\sup_{x\in V}|G(x)|_{2}=\displaystyle\sup_{x\in A}|g(x)|_{2}\quad\mbox{and}\quad\displaystyle\sup_{x_{1}\neq x_{2}\in V}\frac{|G(x_{1})-G(x_{2})|_{2}}{\|x_{1}-x_{2}\|^{\alpha}}=\displaystyle\sup_{x_{1}\neq x_{2}\in A}\frac{|g(x_{1})-g(x_{2})|_{2}}{\|x_{1}-x_{2}\|^{\alpha}}.$$

(B.10)

I. e., satisfying ( B.9 ) on $V$ .

B.5 Lebesgue Measure and Lipschitz Continuous Map

Lemma 6

Let $\emptyset\not=A\subseteq{{\mathbb{R}}^{d}}$ be a Lebesgue-measurable set and $f\colon A\to{{\mathbb{R}}^{d}}$ be Lipschitz continuous. Then

$${\rm meas}\big{(}f(A)\big{)}\leqslant\mathop{\rm Lip}\nolimits_{A}(f)^{d}{\rm meas}(A)$$

(B.11)

and Footnote

Inequality ( B.12 ) is sharp as shown by the example $f=(1+{\delta}){\mathtt{id}}$ .

$$|{\rm meas}(f(A))-{\rm meas}(A)|\leqslant((1+{\delta})^{d}-1){\rm meas}(A),$$

(B.12)

where

$${\delta}:=\mathop{\rm Lip}\nolimits_{A}(f-{\mathtt{id}}).$$

(B.13)

Inequality ( B.12 ) is sharp as shown by the example $f=(1+{\delta}){\mathtt{id}}$ .

Proof

Eq. (B.11) is standart: see, e.g. Theorem 2, Section 2.2 and Theorem 1, Section 2.4 in([ 11 ].

Let us prove (B.13) . By Theorem B.2 , $f-{\mathtt{id}}$ can be extended to a Lipschitz continuous function $g\colon{{\mathbb{R}}^{d}}$ with

$$\mathop{\rm Lip}\nolimits(g)=\mathop{\rm Lip}\nolimits_{A}(f-{\mathtt{id}})={\delta}.$$

By Rademacher’s theorem, there exists a set $N\subseteq{{\mathbb{R}}^{d}}$ with ${\rm meas}(N)=0$ such that $g$ is differentiable on ${{\mathbb{R}}^{d}}\setminus N$ and

$$\|g_{y}\|_{{{\mathbb{R}}^{d}}\setminus N}\leqslant\mathop{\rm Lip}\nolimits_{{{\mathbb{R}}^{d}}\setminus N}(g)\leqslant\mathop{\rm Lip}\nolimits(g)={\delta}.$$

Now pick $y\in{{\mathbb{R}}^{d}}\setminus N$ . Then

$$\displaystyle|\det({\mathsf{1}}_{d}+g_{y}(y))-1|=\left|\int_{0}^{1}\frac{d}{dt}\det({\mathsf{1}}_{d}+tg_{y})dt\right|=\left|\int_{0}^{1}{\rm tr}\left({\rm Adj}({\mathsf{1}}_{d}+tg_{y})g_{y}\right)dt\right|$$

$$\displaystyle\leqslant\int_{0}^{1}d\|{\mathsf{1}}_{d}+tg_{y}\|^{d-1}\|g_{y}\|dt\leqslant\int_{0}^{1}d\left(1+{\delta}t\right)^{d-1}{\delta}dt=(1+{\delta})^{d}-1.$$

Thus, by the change of variable (or area) formula Footnote

See [ 11 ], $\S 3.3$ .

, we have

$$\displaystyle|{\rm meas}(f(A))-{\rm meas}(A)|=|{\rm meas}({({\mathtt{id}}+g)}(A))-{\rm meas}(A)|=\left|\int_{({\mathtt{id}}+g)(A)}dy-\int_{A}dy\right|$$

$$\displaystyle=\left|\int_{({\mathtt{id}}+g)(A\setminus N)}dy-\int_{A\setminus N}dy\right|=\left|\int_{A\setminus N}|\det({\mathsf{1}}_{d}+g_{y})|dy-\int_{A\setminus N}dy\right|$$

$$\displaystyle\leqslant\int_{A\setminus N}|\det({\mathsf{1}}_{d}+g_{y})-1|dy\leqslant((1+{\delta})^{d}-1){\rm meas}(A).$$

$\square$

See [ 11 ], $\S 3.3$ .

B.6 Lipeomorphisms “Close” to Identity

Lemma 7

Let $g\colon{{\mathbb{C}}^{d}}\to{{\mathbb{C}}^{d}}$ be a Lipschitz continuous function such that

$$\displaystyle{\delta}:=\sup_{{{\mathbb{R}}^{d}}}|g-{\mathtt{id}}|<\infty,$$

(B.14)

$$\displaystyle{\theta}:=\mathop{\rm Lip}\nolimits_{{\mathbb{R}}^{d}}(g-{\mathtt{id}})<1.$$

(B.15)

Then $g$ has a Lipschitz global inverse $G$ satisfying

$$\displaystyle\sup_{{{\mathbb{R}}^{d}}}|G-{\mathtt{id}}|\leqslant{\delta},$$

(B.16)

$$\displaystyle\mathop{\rm Lip}\nolimits_{{\mathbb{R}}^{d}}(G-{\mathtt{id}})\|<\frac{1}{1-{\theta}}.$$

(B.17)

Furthermore, for any $\emptyset\neq A\subseteq{{\mathbb{C}}^{d}}$ ,

$$A\subseteq g\left(\overline{{\mathbb{B}}_{{\delta}}(A)}\right).$$

(B.18)

Proof

Let $f:= g-{\mathtt{id}}$ , then, for any $x_{i}\in{{\mathbb{R}}^{d}}$ , one has

$$\displaystyle|g(x_{1})-g(x_{2})|=\big{|}x_{1}-x_{2}+\big{(}f(x_{1})-f(x_{2})\big{)}\big{|}\stackrel{{\scriptstyle(B.15)}}{{>}}|x_{1}-x_{2}|-{\theta}|x_{1}-x_{2}|=(1-{\theta})|x_{1}-x_{2}|,$$

which proves injectivity of $g$ and that

$$\inf_{x_{1}\neq x_{2}}\frac{|g(x_{1})-g(x_{2})|}{|x_{1}-x_{2}|}\geqslant 1-{\theta}>0.$$

(B.19)

Let us now prove (B.18) . Let $\bar{y}\in A$ . It is enough to show that there exists $|y|\leqslant{\delta}$ such that $\bar{y}=g(y+\bar{y})$ , i.e., $y=-f(y+\bar{y})$ , i.e., $y$ is a fixed point of the map

$$h\colon\overline{{\mathbb{B}}_{{\delta}}(0)}\ni y\mapsto-f(y+\bar{y}).$$

But, for any $y\in\overline{{\mathbb{B}}_{{\delta}}(0)}$ ,

$$|h(y)|=|f(y+\bar{y})|\leqslant\|f\|_{{{\mathbb{R}}^{d}}}\stackrel{{\scriptstyle(B.14)}}{{\leqslant}}{\delta},$$

i.e., $h\colon\overline{{\mathbb{B}}_{{\delta}}(0)}\to\overline{{\mathbb{B}}_{{\delta}}(0)}$ . Moreover, $h$ is a contraction since $\mathop{\rm Lip}\nolimits_{\overline{{\mathbb{B}}_{{\delta}}(0)}}(h)\leqslant\mathop{\rm Lip}\nolimits_{{\mathbb{R}}^{d}}(f)\stackrel{{\scriptstyle(B.15)}}{{<}}1$ . Thus, by Banach’s fixed point theorem, we see that (B.18) holds. From (B.18) it follows at once that $g$ is onto ${{\mathbb{R}}^{d}}$ . Now (B.16) and (B.17) follow easily from (B.14) and (B.19) , respectively.     $\square$

B.7 Whitney Smoothness

Definition 1

Let $A\subseteq{{\mathbb{R}}^{d}}$ be non-empty and $n\in{\mathbb{N}}_{0}$ , $m\in{\mathbb{N}}$ . A function $f\colon A\to{\mathbb{R}}^{m}$ is said to be $C^{n}$ on $A$ in the Whitney sense, with Whitney derivatives $(f_{\nu})_{{\nu}\in{\mathbb{N}}_{0}^{d},{|{\nu}|_{1}\leqslant n}}$ , $f_{0}=f$ , and we write $f\in C^{n}_{W}(A,{\mathbb{R}}^{m})$ if, for any ${\varepsilon}>0$ and $y_{0}\in A$ , there exists ${\delta}>0$ such that, for any $y,y^{\prime}\in A\cap{\rm B}_{\delta}(y_{0})$ and ${\nu}\in{\mathbb{N}}_{0}^{d}$ , with ${|{\nu}|_{1}\leqslant n}$ ,

$$\Big{|}f_{\nu}(y^{\prime})-\displaystyle\sum_{\begin{subarray}{c}{\mu}\in{\mathbb{N}}_{0}^{d}\\ {|{\mu}|_{1}\leqslant n-|{\nu}|_{1}}\end{subarray}}\frac{1}{{\mu}!}f_{{\nu}+{\mu}}(y)(y^{\prime}-y)^{\mu}\Big{|}\leqslant{\varepsilon}|y^{\prime}-y|^{n-|{\nu}|_{1}}.$$

(B.20)

Lemma 8 ( [ 8 ] ; [ 12 ] )

Let $A\subseteq{{\mathbb{R}}^{d}}$ be non-empty and $n\in{\mathbb{N}}_{0}$ . For $m\in{\mathbb{N}}$ , let $f_{m}$ be a real-analytic function with holomorphic extension to $D_{r_{m}}(A)$ , with $r_{m}\downarrow 0$ as $m\rightarrow\infty$ . Assume that

$$a:=\displaystyle\sum_{m=1}^{\infty}\|f_{m}\|_{r_{m},A}r_{m}^{-n}<\infty,\qquad\|f_{m}\|_{r_{m},A}:=\displaystyle\sup_{{\rm B}^{d}_{r_{m}}(A)}|f_{m}|.$$

(B.21)

Then $f:=\displaystyle\sum_{m=1}^{\infty}f_{m}\in C^{n}_{W}(A,{\mathbb{R}})$ with Whitney derivatives $f_{\nu}:=\displaystyle\sum_{m=1}^{\infty}\partial_{y}^{\nu}f_{m}$ .

For completeness, we recall the beautiful Whitney extension theorem.

Theorem B.3 ( [ 19 ] )

Let $A\subseteq{{\mathbb{R}}^{d}}$ be a closed set and $f\in C^{n}_{W}(A,{\mathbb{R}})$ , $n\in{\mathbb{N}}_{0}$ . Then there exists $\bar{f}\in C^{n}({{\mathbb{R}}^{d}},{\mathbb{R}})$ , real-analytic on ${{\mathbb{R}}^{d}}\setminus A$ and such that $D^{\nu}\bar{f}=f_{\nu}$ on $A$ , for any ${\nu}\in{\mathbb{N}}_{0}^{d}$ , with ${|{\nu}|_{1}\leqslant n}$ .

B.8 Measure of Tubular Neighbourhoods of Hypersurfaces

Recall the definitions of minimal focal distance and of inner domains given in Section 3.2 . The first elementary remark is that, for smooth domains, taking ${\rho}$ –inner domains is the inverse operation of taking ${\rho}$ -neighbourhood:

Lemma 9

Let ${\mathscr{D}}\subseteq{{\mathbb{R}}^{d}}$ be an open and bounded set with $C^{2}$ boundary $\partial{\mathscr{D}}=S$ compact and connected. Then, for any $0<{\rho}^{\prime}<{\rho}\leqslant{\rm minfoc}(S)$ , one has

$${\bf B}_{\rho}\big{(}{\mathscr{D}}^{\prime\prime}_{\rho}\big{)}={\mathscr{D}},\qquad\mbox{and}\qquad{\bf B}_{{\rho}-{\rho}^{\prime}}\big{(}{\mathscr{D}}^{\prime\prime}_{\rho}\big{)}={\mathscr{D}}^{\prime\prime}_{{\rho}^{\prime}}.$$

(B.22)

Proof

We start proving the first part of (B.22) . By definition, ${\bf B}_{\rho}\big{(}{\mathscr{D}}^{\prime\prime}_{\rho}\big{)}\subseteq{\mathscr{D}}$ . Thus, it remains only to show that ${\mathscr{D}}\setminus{\mathscr{D}}^{\prime\prime}_{\rho}\subseteq{\bf B}_{\rho}\big{(}{\mathscr{D}}^{\prime\prime}_{\rho}\big{)}$ . Let then $y_{0}\in{\mathscr{D}}\setminus{\mathscr{D}}^{\prime\prime}_{\rho}$ . As $S$ is compact and $\mathop{\rm dist}\nolimits_{2}$ is continuous, there exists $\bar{y}_{0}\in S$ such that $\mathop{\rm dist}\nolimits_{2}(y_{0},{{\mathbb{R}}^{d}}\setminus{\mathscr{D}}\big{)}=\mathop{\rm dist}\nolimits_{2}(y_{0},S\big{)}=|y_{0}-\bar{y}_{0}|_{2}$ . The vector ${\nu}:=(y_{0}-\bar{y}_{0})/|y_{0}-\bar{y}_{0}|_{2}$ is the inward unit normal to $\partial{\mathscr{D}}=S$ at $\bar{y}_{0}$ . Indeed, for any smooth curve ${\gamma}\colon[0,1]\to S$ with ${\gamma}(0)=\bar{y}_{0}$ , $0$ is a minimum of the smooth map $f(t):=|{\gamma}(t)-y_{0}|_{2}^{2}$ . Thus,

$$0=f^{\prime}(0)=2\dot{{\gamma}}(0)\cdot(\bar{y}_{0}-y_{0}).$$

which, by the arbitrariness of ${\gamma}$ , implies that the line $(\bar{y}_{0}y_{0})$ is perpendicular to the tangent space to $S$ at $\bar{y}_{0}$ and, therefore ${\nu}$ is the inward unit normal to $\partial{\mathscr{D}}$ at $\bar{y}_{0}$ . Let $y_{1}:=\bar{y}_{0}+{\rho}{\nu}$ . By assumption, we have $\mathop{\rm dist}\nolimits_{2}(y_{1},S\big{)}={\rho}$ , and, therefore, $y_{1}\in{\mathscr{D}}$ . In addition, $y_{1}\in{\mathscr{D}}^{\prime\prime}_{\rho}$ . Indeed, for any $y\in{\bf B}_{\rho}(y_{1})$ , $\mathop{\rm dist}\nolimits_{2}\bigl{(}y,{{\mathbb{R}}^{d}}\setminus{\mathscr{D}}\bigr{)}\geqslant\mathop{\rm dist}\nolimits_{2}\bigl{(}y_{1},{{\mathbb{R}}^{d}}\setminus{\mathscr{D}}\bigr{)}-|y_{1}-y|_{2}=\mathop{\rm dist}\nolimits_{2}\bigl{(}y_{1},S\bigr{)}-|y_{1}-y|_{2}={\rho}-|y_{1}-y|_{2}>0$ . Thus, as ${{\mathbb{R}}^{d}}\setminus{\mathscr{D}}$ is a closed set, $y\not\in{{\mathbb{R}}^{d}}\setminus{\mathscr{D}}$ , i. e., $y\in{\mathscr{D}}$ . Hence, ${\bf B}_{\rho}(y_{1})\subseteq{\mathscr{D}}$ , i.e., $y_{1}\in{\mathscr{D}}^{\prime\prime}_{\rho}$ . In particular, the argument above shows that: Footnote

Actually, one checks easily that $\partial{\mathscr{D}}^{\prime\prime}_{\rho}=\{y\in{{\mathbb{R}}^{d}}:\mathop{\rm dist}\nolimits_{2}\bigl{(}y,{{\mathbb{R}}^{d}}\setminus{\mathscr{D}}\big{)}={\rho}\}$ and ${\rm int}({\mathscr{D}}^{\prime\prime}_{\rho})=\{y\in{{\mathbb{R}}^{d}}:\mathop{\rm dist}\nolimits_{2}\bigl{(}y,{{\mathbb{R}}^{d}}\setminus{\mathscr{D}}\big{)}>{\rho}\}$ , ${\rm int}({\mathscr{D}}^{\prime\prime}_{\rho})$ being the interior of ${\mathscr{D}}^{\prime\prime}_{\rho}$ .

for any $y\in{{\mathbb{R}}^{d}}$ , $\mathop{\rm dist}\nolimits_{2}\bigl{(}y,{{\mathbb{R}}^{d}}\setminus{\mathscr{D}}\big{)}\geqslant{\rho}$ implies that $y\in{\mathscr{D}}^{\prime\prime}_{\rho}$ . Thus, as $y_{0}\in{\mathscr{D}}\setminus{\mathscr{D}}^{\prime\prime}_{\rho}$ , we have $\mathop{\rm dist}\nolimits_{2}\bigl{(}y_{0},{{\mathbb{R}}^{d}}\setminus{\mathscr{D}}\big{)}<{\rho}$ , which means $y_{0}$ is in the open segment $(\bar{y}_{0},y_{1})$ . Therefore, $|y_{0}-y_{1}|_{2}<|\bar{y}_{0}-y_{1}|_{2}={\rho}$ , i.e., $y_{0}\in{\bf B}_{\rho}(y_{1})\subseteq{\bf B}_{\rho}({\mathscr{D}}^{\prime\prime}_{\rho})$ . We now prove the second part of (B.22) . We have ${\bf B}_{{\rho}-{\rho}^{\prime}}\big{(}{\mathscr{D}}^{\prime\prime}_{\rho}\big{)}\subseteq{\mathscr{D}}^{\prime\prime}_{{\rho}^{\prime}}$ . Indeed, for any $y_{0}\in{\mathscr{D}}^{\prime\prime}_{{\rho}}$ , $y_{1}\in{\bf B}_{{\rho}-{\rho}^{\prime}}(y_{0})$ and $y\in{\bf B}_{{\rho}^{\prime}}(y_{1})$ ,

$$|y-y_{0}|\leqslant|y-y_{1}|+|y_{1}-y_{0}|<{\rho}^{\prime}+({\rho}-{\rho}^{\prime})={\rho}\quad i.e.,\ y\in{\bf B}_{{\rho}}(y_{0}),$$

which implies ${\bf B}_{{\rho}-{\rho}^{\prime}}\big{(}{\mathscr{D}}^{\prime\prime}_{\rho}\big{)}\subseteq{\mathscr{D}}^{\prime\prime}_{{\rho}^{\prime}}$ . It remains to show that ${\mathscr{D}}^{\prime\prime}_{{\rho}^{\prime}}{\backslash}{\mathscr{D}}^{\prime\prime}_{{\rho}}\subseteq{\bf B}_{{\rho}-{\rho}^{\prime}}\big{(}{\mathscr{D}}^{\prime\prime}_{\rho}\big{)}$ . The proof follows analogously to the previous one. Let $y_{0}\in{\mathscr{D}}^{\prime\prime}_{{\rho}^{\prime}}{\backslash}{\mathscr{D}}^{\prime\prime}_{{\rho}}$ and $\bar{y}_{0}\in S$ such that $\mathop{\rm dist}\nolimits_{2}(y_{0},{{\mathbb{R}}^{d}}\setminus{\mathscr{D}}\big{)}=\mathop{\rm dist}\nolimits_{2}(y_{0},S\big{)}=|y_{0}-\bar{y}_{0}|_{2}$ . Then ${\rho}^{\prime}\leqslant|y_{0}-\bar{y}_{0}|_{2}<{\rho}$ , and the vector ${\nu}:=(y_{0}-\bar{y}_{0})/|y_{0}-\bar{y}_{0}|_{2}$ is the inward unit normal to $\partial{\mathscr{D}}=S$ at $\bar{y}_{0}$ . Set $y_{1}^{\prime}:=\bar{y}_{0}+{\rho}^{\prime}{\nu}$ . Thus, $|y_{1}^{\prime}-\bar{y}_{0}|_{2}={\rho}^{\prime}\leqslant|y_{0}-\bar{y}_{0}|_{2}$ and, hence, $y_{1}^{\prime}\in{\mathscr{D}}^{\prime\prime}_{{\rho}^{\prime}}$ and $y_{1}^{\prime}$ is in the semi-open segment $(\bar{y}_{0},y_{0}]$ . Therefore, $|y_{1}^{\prime}-{y}_{0}|_{2}=|y_{0}-\bar{y}_{0}|_{2}-|y_{1}^{\prime}-\bar{y}_{0}|_{2}<{\rho}-{\rho}^{\prime}$ . Hence, $y_{0}\in{\bf B}_{{\rho}-{\rho}^{\prime}}(y_{1}^{\prime})\subseteq{\bf B}_{{\rho}-{\rho}^{\prime}}({\mathscr{D}}^{\prime\prime}_{{\rho}^{\prime}})$ , i.e., ${\mathscr{D}}^{\prime\prime}_{{\rho}^{\prime}}{\backslash}{\mathscr{D}}^{\prime\prime}_{{\rho}}\subseteq{\bf B}_{{\rho}-{\rho}^{\prime}}\big{(}{\mathscr{D}}^{\prime\prime}_{\rho}\big{)}$ .     $\square$

Actually, one checks easily that $\partial{\mathscr{D}}^{\prime\prime}_{\rho}=\{y\in{{\mathbb{R}}^{d}}:\mathop{\rm dist}\nolimits_{2}\bigl{(}y,{{\mathbb{R}}^{d}}\setminus{\mathscr{D}}\big{)}={\rho}\}$ and ${\rm int}({\mathscr{D}}^{\prime\prime}_{\rho})=\{y\in{{\mathbb{R}}^{d}}:\mathop{\rm dist}\nolimits_{2}\bigl{(}y,{{\mathbb{R}}^{d}}\setminus{\mathscr{D}}\big{)}>{\rho}\}$ , ${\rm int}({\mathscr{D}}^{\prime\prime}_{\rho})$ being the interior of ${\mathscr{D}}^{\prime\prime}_{\rho}$ .

The next result gives a precise evaluation of tubular domains in the case where the metric is the Euclidean one . Define

$${\mathfrak{T}}_{{\rho}}(S):=\{u\in{{\mathbb{R}}^{d}}:\ \mathop{\rm dist}\nolimits_{2}(u,S)<{\rho}\}.$$

(B.23)

Lemma 10

Let ${\mathscr{D}}\subseteq{{\mathbb{R}}^{d}}$ be a bounded set with $C^{2}$ boundary $\partial{\mathscr{D}}=S$ compact and connected. Then, for any $0<{\rho}\leqslant{\rm minfoc}(S)$ , then,

$$\displaystyle{\rm meas}({\mathfrak{T}}_{{\rho}}(S))\leqslant{\frac{2}{d}}\frac{(1+{\rho}\kappa)^{d}-1}{\kappa}\mathcal{H}^{d-1}(S),$$

(B.24)

where $\kappa:=\sup_{S}\max_{1\leqslant j\leqslant d-1}|\kappa_{j}|$ with $\kappa_{j}$ the principal curvatures of $S$ , while $\mathcal{H}^{d-1}$ denotes the $(d-1)$ -dimensional Hausdorff measure ( “surface area” ) .

Proof

Footnote

Compare [ 18 ], Ch. 1.

Compare [ 18 ], Ch. 1.

We will estimate the “inner tubular neighbourhoods”

$${\mathfrak{T}}_{\rho}^{\prime}(S):=\{y\in{\mathscr{D}}:\mathop{\rm dist}\nolimits_{2}(y,S)<{\rho}\},$$

as the argument for “outer tubular neighbourhood” $\{y\notin{\mathscr{D}}:\mathop{\rm dist}\nolimits_{2}(y,S)<{\rho}\}$ is completely analogous.

Since $S$ is compact and connected, we may assume that $S=f^{-1}(\{0\})$ with $f\in C^{2}({\mathbb{R}}^{d},{\mathbb{R}})$ and $0$ a regular value for $f$ . Set

$$\nu(x)=\frac{\nabla f}{|\nabla f|_{2}},\qquad|\cdot|_{2}:=\mathop{\rm dist}\nolimits_{2}(\cdot,0),$$

and replacing eventually $f$ by $-f$ , we can assume that $\nu$ is the inwards unit normal vector field of $S$ . Let $\{\phi_{j}\colon U_{j}\to{\mathbb{R}}^{m}\}_{j=1}^{p}$ be an atlas of $S$ ,

$$\Psi_{j}(u,t):=\phi_{j}(u)+t\nu(\phi_{j}(u)),\qquad O_{j}:=\Psi_{j}(U_{j}\times[0,{\rho})),$$

and observe that ²⁹

$${\mathfrak{T}}_{{\rho}}^{\prime}(S)=\bigcup_{j=1}^{p}O_{j}.$$

Let $\{\psi_{j}\}_{j=1}^{p}$ be a partition of unity subordinated to the open covering of $\{{O}_{j}\}_{j=1}^{p}$ of ${\mathfrak{T}}^{\prime}_{{\rho}}(S)$ , i. e.,

• $\psi_{j}\in C^{\infty}_{c}({\mathfrak{T}}^{\prime}_{{\rho}}(S))$ ;

• $0\leqslant\psi_{j}\leqslant 1$ ;

• ${\rm supp}\psi_{j}\subseteq{O}_{j}$ ;

• $\displaystyle\sum_{j=1}^{p}\psi_{j}\equiv 1$ on ${\mathfrak{T}}^{\prime}_{{\rho}}(S)$ .

Given $1\leqslant j\leqslant p$ , define $n_{j}\colon U_{j}\longrightarrow\mathbb{S}^{d}=\{x\in{\mathbb{R}}^{d}:|x|_{2}=x_{1}^{2}+\cdots+x_{d}^{2}=1\}\subseteq{\mathbb{R}}^{d}$ as

$$n_{j}:=\nu\circ\phi_{j},$$

and $K_{j}\colon U_{j}\longrightarrow T^{*}S$ such that ³⁰

$$K_{j}(u):=-\nu^{\prime}(\phi_{j}(u)).$$

Then $K_{j}$ is symmetric ³¹ and therefore diagonalizable, with eigenvalues $\kappa_{i}\circ\phi_{j}^{-1}$ , $1\leqslant i\leqslant d-1$ , and satisfies

$${\frac{\partial n_{j}}{\partial u}}=-K_{j}\frac{\partial\phi_{j}}{\partial u}.$$

(B.25)

Thus, recalling that $0=\partial_{x}\nu^{2}=2\nu^{\prime}\cdot\nu$ , we have

$$\displaystyle{\rm meas}({\mathfrak{T}}^{\prime}_{\rho}(S))=\sum_{j=1}^{p}\int_{{O}_{j}}\psi_{j}dudt$$

$$\displaystyle=\sum_{j=1}^{p}\int_{\Psi_{j}(U_{j}\times[0,{\rho}))}\psi_{j}dudt$$

$$\displaystyle=\sum_{j=1}^{p}\int_{U_{j}\times[0,{\rho})}\Psi_{j}^{*}(\psi_{j}dudt)$$

$$\displaystyle=\sum_{j=1}^{p}\int_{U_{j}\times[0,{\delta})}\psi_{j}\circ\Psi_{j}\left|\det\left(\frac{\partial\Psi_{j}}{\partial(u,t)}\right)\right|dudt$$

$$\displaystyle\stackrel{{\scriptstyle(B.25)}}{{=}}\sum_{j=1}^{p}\int_{U_{j}\times[0,{\rho})}\psi_{j}\circ\Psi_{j}\left|\det\left[\frac{\partial\phi_{j}}{\partial u}-tK_{j}\cdot\frac{\partial\phi_{j}}{\partial u},\nu(\phi_{j}(u))\right]\right|dudt$$

$$\displaystyle=\sum_{j=1}^{p}\int_{U_{j}\times[0,{\rho})}\psi_{j}\circ\Psi_{j}\left|\det\left(\begin{pmatrix}{\mathsf{1}}_{d-1}-tK_{j}\end{pmatrix}\left[\frac{\partial\phi_{j}}{\partial u},\nu(\phi_{j}(u))\right]\right)\right|dudt$$

$$\displaystyle=\sum_{j=1}^{p}\int_{U_{j}\times[0,{\rho})}\psi_{j}\circ\Psi_{j}\left|\det({\mathsf{1}}_{d-1}-tK_{j})\right|\left|\det\left[\frac{\partial\phi_{j}}{\partial u},\nu(\phi_{j}(u))\right]\right|dudt$$

$$\displaystyle\leqslant\int_{0}^{\rho}\sum_{j=1}^{p}\int_{U_{j}}\psi_{j}\big{(}\phi_{j}(u)+t\nu(\phi_{j}(u))\big{)}\left|\det\left[\frac{\partial\phi_{j}}{\partial u},\nu(\phi_{j}(u))\right]\right|du(1+t\kappa)^{d-1}dt$$

$$\displaystyle{=\int_{0}^{\rho}\sum_{j=1}^{p}\int_{U_{j}}\psi_{j}\big{(}\phi_{j}(u)+t\nu(\phi_{j}(u))\big{)}\left(\det\left(\frac{\partial\phi_{j}}{\partial u}\right)^{T}\frac{\partial\phi_{j}}{\partial u}\right)^{1/2}du(1+t\kappa)^{d-1}dt}$$

$$\displaystyle{=\int_{0}^{\rho}\sum_{j=1}^{p}\int_{\phi_{j}(U_{j})}\psi_{j}\big{(}x+t\nu(x)\big{)}d\mathcal{H}^{d-1}(x)(1+t\kappa)^{d-1}dt}\quad\mbox{\footnotesize(see [Theorem 2, pg. 99])}$$

$$\displaystyle\stackrel{{\scriptstyle\mathbf{(ii)}}}{{\leqslant}}\int_{0}^{\rho}\sum_{j=1}^{p}\int_{\bigcup_{i=1}^{p}\phi_{i}(U_{i})}\psi_{j}\big{(}x+t\nu(x)\big{)}d\mathcal{H}^{d-1}(x)(1+t\kappa)^{d-1}dt$$

$$\displaystyle=\int_{0}^{\rho}\int_{S}\sum_{j=1}^{p}\psi_{j}\big{(}x+t\nu(x)\big{)}d\mathcal{H}^{d-1}(x)(1+t\kappa)^{d-1}dt$$

$$\displaystyle\stackrel{{\scriptstyle\mathbf{(iv)}}}{{=}}\int_{0}^{\rho}\int_{S}d\mathcal{H}^{d-1}(x)(1+t\kappa)^{d-1}dt$$

$$\displaystyle=\frac{(1+{\rho}\kappa)^{d}-1}{d\kappa}\mathcal{H}^{d-1}(S).$$

$\square$

B.9 Kolmogorov Non-degenerate Normal Forms

Let $H\colon\mathcal{M}:={{\mathbb{R}}^{d}}\times{{\mathbb{T}}^{d}}\to{\mathbb{R}}$ be a $C^{2}$ -Hamiltonian. An embedded torus $\mathcal{T}$ in $\mathcal{M}$ is said to be $H$ -Kolmogorov non-degenerate if there exists a neighbourhood $\mathcal{M}_{0}$ of $\{0\}\times{{\mathbb{T}}^{d}}$ in $\mathcal{M}$ , a symplectic change of coordinates $\phi\colon\mathcal{M}_{0}\to\mathcal{M}$ with $\phi(\{0\}\times{{\mathbb{T}}^{d}})=\mathcal{T}$ , a constant $E\in{\mathbb{R}}$ , a vector ${\omega}\in{{\mathbb{R}}^{d}}$ and a function $Q\colon\mathcal{M}_{0}\to{\mathbb{R}}$ of class $C^{2}$ such that

$$H\circ\phi(y,x)=E+{\omega}\cdot y+Q(y,x)\qquad\mbox{and}\qquad\partial_{y}^{\mu}Q(0,\cdot)\equiv 0,\ \forall\mu\in{\mathbb{N}}_{0}^{d},\ |\mu|_{1}\leqslant 1,$$

(B.26)

and

$$\det{\left\langle\partial_{yy}Q(0,\cdot)\right\rangle}\not=0.$$

(B.27)

A Hamiltonian $H$ in the form (B.26) is said to be in Kolmogorov normal form. The Kolmogorov normal form is said to be non-degenerate if, in addition, the quadratic (in $y$ ) part $Q$ satisfies (B.27) .