height=2.92223pt,depth=4.25003pt}}_{\,\\{0,1\\}}}=0$ for $0\leqslant
l\leqslant k-1$. Let $\varphi\in H^{2k+2}_{(0)}(0,1)$. Then,

  1. 1.

$A\varphi\in H^{2k}_{(0)}(0,1)$, so
$\underline{\operatorname{ad}}_{A}^{k}(B_{1})(A\varphi)$ is well-defined, by
induction.

  2. 2.

We now prove that $\underline{\operatorname{ad}}_{A}^{k}(B_{1})\varphi\in
H^{2}_{(0)}(0,1)$. The induction hypothesis gives:
$\underline{\operatorname{ad}}_{A}^{k}(B_{1})\varphi=\displaystyle\sum_{l=0}^{k}\alpha_{l}^{k}\mu_{1}^{(2k-l)}\varphi^{(l)}.$
Then, one has $\underline{\operatorname{ad}}_{A}^{k}(B_{1})\in H^{2}(0,1)$.
Moreover, for all $l\in\llbracket 0,k\rrbracket$,

    1. a.

If $l$ is odd, $l=2i+1$, and
$\mathchoice{{\mu_{1}^{(2k-2i-1)}\varphi^{(l)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu_{1}^{(2k-2i-1)}\varphi^{(l)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu_{1}^{(2k-2i-1)}\varphi^{(l)}\,\smash{\vrule
height=3.61111pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu_{1}^{(2k-2i-1)}\varphi^{(l)}\,\smash{\vrule
height=2.92223pt,depth=4.25003pt}}_{\,\\{0,1\\}}}=0$ because $2k-2i-1\leqslant
2k-1$ so $\mathchoice{{\mu_{1}^{(2k-2i-1)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu_{1}^{(2k-2i-1)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu_{1}^{(2k-2i-1)}\,\smash{\vrule
height=3.61111pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu_{1}^{(2k-2i-1)}\,\smash{\vrule
height=2.92223pt,depth=4.25003pt}}_{\,\\{0,1\\}}}=0$.

    2. b.

Otherwise, $l$ is even,
$\mathchoice{{\mu_{1}^{(2k-l)}\varphi^{(l)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu_{1}^{(2k-l)}\varphi^{(l)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu_{1}^{(2k-l)}\varphi^{(l)}\,\smash{\vrule
height=3.61111pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu_{1}^{(2k-l)}\varphi^{(l)}\,\smash{\vrule
height=2.92223pt,depth=4.25003pt}}_{\,\\{0,1\\}}}=0$ because $l\leqslant k$
and $\varphi\in H^{2k+2}_{(0)}(0,1)$.

Then, the bracket is well-defined. The proposition is true by induction on
$k$.

∎

######  Proposition A.6.

Let $k\in\mathbb{N}^{*}$, $\mu_{1}\in W^{4k-2,\infty}(0,1)$ with all its
derivatives of odd order less than or equal to $4k-5$ that vanish at the
boundary. Then, for all $\varphi\in H^{3k-1}_{(0)}(0,1)$ the operator
$[\underline{\operatorname{ad}}_{A}^{k-1}(B_{1}),\underline{\operatorname{ad}}_{A}^{k}(B_{1})]\varphi$
is well-defined in $L^{2}(0,1)$.

###### Proof.

Let $\varphi\in H^{3k-1}_{(0)}(0,1)$. Then $\varphi\in H^{2k}_{(0)}(0,1)$ so
$\underline{\operatorname{ad}}_{A}^{k}(B_{1})\varphi$ and
$\underline{\operatorname{ad}}_{A}^{k-1}(B_{1})\varphi$ are well-defined by
Proposition A.5 and one has to prove that:

  1. 1.

$\underline{\operatorname{ad}}_{A}^{k-1}(B_{1})\varphi\in H^{2k}_{(0)}(0,1)$.
One recalls:
$\underline{\operatorname{ad}}_{A}^{k-1}(B_{1})\varphi=\displaystyle\sum_{l=0}^{k-1}\alpha_{l}^{k-1}\mu_{1}^{(2k-l-2)}\varphi^{(l)}$.
According to the regularity of $\varphi$ and $\mu_{1}$, this expansion gives
$\underline{\operatorname{ad}}_{A}^{k-1}(B_{1})\varphi\in H^{2k}(0,1)$. Then,
using Leibniz formula, for all $j\in\llbracket 0,k-1\rrbracket$,
$l\in\llbracket 0,k-1\rrbracket$,

$\left(\mu_{1}^{(2k-l-2)}\varphi^{(l)}\right)^{(2j)}=\displaystyle\sum_{i=0}^{2j}\binom{i}{2j}\mu_{1}^{(2k-l-2+i)}\varphi^{(l+2j-i)}.$

Each term of this sum vanish at $x=0,1$. Indeed,

    1. a.

If $i$ and $l$ have the same parity, $l+2j-i$ is even and $l+2j-i\leqslant
k-1+2(k-1)=3k-3$ so $\mathchoice{{\varphi^{(l+2j-i)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\varphi^{(l+2j-i)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\varphi^{(l+2j-i)}\,\smash{\vrule
height=3.61111pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\varphi^{(l+2j-i)}\,\smash{\vrule
height=2.92223pt,depth=4.25003pt}}_{\,\\{0,1\\}}}=0$.

    2. b.

Otherwise, $2k-l-2+i$ is odd and $2k-l-2+i\leqslant 2k-2+2j-1\leqslant 4k-5$
so $\mathchoice{{\mu_{1}^{(2k-l+i)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu_{1}^{(2k-l+i)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu_{1}^{(2k-l+i)}\,\smash{\vrule
height=3.61111pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu_{1}^{(2k-l+i)}\,\smash{\vrule
height=2.92223pt,depth=4.25003pt}}_{\,\\{0,1\\}}}=0$.

  2. 2.

$\underline{\operatorname{ad}}_{A}^{k}(B_{1})\varphi\in H^{2k-2}_{(0)}(0,1)$.
This is proven in the same way as point $1$.

Consequently, the operator
$\left[\underline{\operatorname{ad}}_{A}^{k-1}(B_{1}),\underline{\operatorname{ad}}_{A}^{k}(B_{1})\right]$
is well-defined. ∎

Then, if $2\lfloor\frac{n+1}{2}\rfloor-3\leqslant p-1$, Proposition A.6
ensures that the brackets in (1.7) and (1.8) are well-defined.

######  Proposition A.7.

Let $k\in\mathbb{N}$, $\mu_{1},\mu_{2}\in W^{2k,\infty}(0,1)$ with all their
derivatives of odd order less than or equal to $2k-3$ that vanish at the
boundary Then, for all $\varphi\in
H^{k+\lfloor\frac{k+1}{2}\rfloor}_{(0)}(0,1)$,
$\left[\underline{\operatorname{ad}}_{A}^{\lfloor\frac{k+1}{2}\rfloor}(B_{1}),\underline{\operatorname{ad}}_{A}^{\lfloor\frac{k}{2}\rfloor}(B_{2})\right]ph$
is well-defined in $L^{2}(0,1)$.

This proposition can be demonstrated in the same way as the previous one.
Similarly, if $n-2\leqslant p-1$, the brackets in (1.9) are well-defined,
thanks Proposition A.7.

####  A.2.2 Computation of Lie brackets for the Schrödinger equation

######  Lemma A.8.

Let $k\in\mathbb{N}$ and $\mu\in H^{2k}(0,1)$ be such that
$\mathchoice{{\mu^{(2l+1)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu^{(2l+1)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu^{(2l+1)}\,\smash{\vrule
height=3.61111pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{\mu^{(2l+1)}\,\smash{\vrule
height=2.92223pt,depth=4.25003pt}}_{\,\\{0,1\\}}}=0$ for $0\leqslant
l\leqslant k-2$. Then, for all $p,q\in\mathbb{N}^{*}$,

$\left\langle\underline{\operatorname{ad}}_{A}^{k}(\mu)\varphi_{p},\varphi_{q}\right\rangle=(\lambda_{p}-\lambda_{q})^{k}\left\langle\mu\varphi_{p},\varphi_{q}\right\rangle.$

###### Proof.

If $k=0$, the result in true. Then, assume that $k\geqslant 1$. As
$\varphi_{p},\varphi_{q}\in H^{2k}_{(0)}(0,1)$, one has:

$\left\langle\underline{\operatorname{ad}}_{A}^{k}(\mu)\varphi_{p},\varphi_{q}\right\rangle=\left\langle[\underline{\operatorname{ad}}_{A}^{k-1}(\mu),A]\varphi_{p},\varphi_{q}\right\rangle=\left\langle(\lambda_{p}I_{d}-A)\underline{\operatorname{ad}}_{A}^{k-1}(\mu)\varphi_{p},\varphi_{q}\right\rangle$

Using the symmetry property of $A$,

$\left\langle\underline{\operatorname{ad}}_{A}^{k}(\mu)\varphi_{p},\varphi_{q}\right\rangle=\left\langle\underline{\operatorname{ad}}_{A}^{k-1}(\mu)\varphi_{p},(\lambda_{p}I_{d}-A)\varphi_{q}\right\rangle=(\lambda_{p}-\lambda_{q})\left\langle\underline{\operatorname{ad}}_{A}^{k-1}(\mu)\varphi_{p},\varphi_{q}\right\rangle.$

An induction gives the result. ∎

######  Proposition A.9.

Let $k\in\mathbb{N}^{*}$ be a fixed integer, $K\geqslant 2$ and $\mu_{1}\in
H^{4k-2}(0,1)$ be such that, for $0\leqslant l\leqslant 2k-3$
$\mathchoice{{{\mu_{1}}^{(2l+1)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{{\mu_{1}}^{(2l+1)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{{\mu_{1}}^{(2l+1)}\,\smash{\vrule
height=3.61111pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{{\mu_{1}}^{(2l+1)}\,\smash{\vrule
height=2.92223pt,depth=4.25003pt}}_{\,\\{0,1\\}}}=0$. Then,

$\left\langle[\underline{\operatorname{ad}}_{A}^{k-1}(\mu_{1}),\underline{\operatorname{ad}}_{A}^{k}(\mu_{1})]\varphi_{1},\varphi_{K}\right\rangle=2(-1)^{k}A_{k}^{1}.$

###### Proof.

By definition,

$2(-1)^{k}A_{k}^{1}=\sum_{j=1}^{+\infty}c_{j}(\lambda_{j}-\lambda_{1})^{k-1}(\lambda_{K}-\lambda_{j})^{k}-\sum_{j=1}^{+\infty}c_{j}(\lambda_{j}-\lambda_{1})^{k}(\lambda_{K}-\lambda_{j})^{k-1}.$

Using Lemma A.8, we obtain:

$\sum_{j=1}^{+\infty}\left(\left\langle\underline{\operatorname{ad}}_{A}^{k-1}(\mu_{1})\varphi_{j},\varphi_{1}\right\rangle\left\langle\underline{\operatorname{ad}}_{A}^{k}(\mu_{1})\varphi_{K},\varphi_{j}\right\rangle-\left\langle\underline{\operatorname{ad}}_{A}^{k}(\mu_{1})\varphi_{j},\varphi_{1}\right\rangle\left\langle\underline{\operatorname{ad}}_{A}^{k-1}(\mu_{1})\varphi_{K},\varphi_{j}\right\rangle\right).$

Using the symmetry of the operators, we get:

$2(-1)^{k}A_{k}^{1}=(-1)^{k-1}\left\langle\underline{\operatorname{ad}}_{A}^{k-1}(\mu_{1})\varphi_{1},\underline{\operatorname{ad}}_{A}^{k}(\mu_{1})\varphi_{K}\right\rangle+(-1)^{k-1}\left\langle\underline{\operatorname{ad}}_{A}^{k}(\mu_{1})\varphi_{1},\underline{\operatorname{ad}}_{A}^{k-1}(\mu_{1})\varphi_{K}\right\rangle.$

Then,

$2(-1)^{k}A_{k}^{1}=-\left\langle\underline{\operatorname{ad}}_{A}^{k}(\mu_{1})\left(\underline{\operatorname{ad}}_{A}^{k-1}(\mu_{1})\varphi_{1}\right),\varphi_{K}\right\rangle+\left\langle\underline{\operatorname{ad}}_{A}^{k-1}(\mu_{1})\left(\underline{\operatorname{ad}}_{A}^{k}(\mu_{1})\varphi_{1}\right),\varphi_{K}\right\rangle.$

Finally,

$2(-1)^{k}A_{k}^{1}=\left\langle[\underline{\operatorname{ad}}_{A}^{k-1}(\mu_{1}),\underline{\operatorname{ad}}_{A}^{k}(\mu_{1})]\varphi_{1},\varphi_{K}\right\rangle.$

∎

######  Proposition A.10.

Let $k\in\mathbb{N}$, $K\geqslant 2$, and $\mu_{1},\mu_{2}\in H^{2k}(0,1)$
such that $\mathchoice{{{\mu_{i}}^{(2l+1)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{{\mu_{i}}^{(2l+1)}\,\smash{\vrule
height=5.12445pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{{\mu_{i}}^{(2l+1)}\,\smash{\vrule
height=3.61111pt,depth=4.25003pt}}_{\,\\{0,1\\}}}{{{\mu_{i}}^{(2l+1)}\,\smash{\vrule
height=2.92223pt,depth=4.25003pt}}_{\,\\{0,1\\}}}=0$ for $0\leqslant
l\leqslant k-2$ and $1\leqslant i\leqslant 2$. Then,

$\left\langle\left[\underline{\operatorname{ad}}_{A}^{\lfloor\frac{k+1}{2}\rfloor}(\mu_{1}),\underline{\operatorname{ad}}_{A}^{\lfloor\frac{k}{2}\rfloor}(\mu_{2})\right]\varphi_{1},\varphi_{K}\right\rangle=(-1)^{k}\gamma_{k}.$

###### Proof.

Using Lemma A.8, we notice that

$\gamma_{k}=\displaystyle\sum_{j=1}^{+\infty}\left(\left\langle\underline{\operatorname{ad}}_{A}^{\lfloor\frac{k+1}{2}\rfloor}(\mu_{1})\varphi_{K},\varphi_{j}\right\rangle\left\langle\underline{\operatorname{ad}}_{A}^{\lfloor\frac{k}{2}\rfloor}(\mu_{2})\varphi_{j},\varphi_{1}\right\rangle\right.-\\\
\left.\left\langle\underline{\operatorname{ad}}_{A}^{\lfloor\frac{k}{2}\rfloor}(\mu_{2})\varphi_{K},\varphi_{j}\right\rangle\left\langle\underline{\operatorname{ad}}_{A}^{\lfloor\frac{k+1}{2}\rfloor}(\mu_{1})\varphi_{j},\varphi_{1}\right\rangle\right)$

As $(\varphi_{j})_{j\geqslant 1}$ is an orthogonal basis of $L^{2}$, and
thanks to the symmetry/skew-symmetry of the operator,

$\gamma_{k}=(-1)^{\lfloor\frac{k}{2}\rfloor}\left\langle\underline{\operatorname{ad}}_{A}^{\lfloor\frac{k+1}{2}\rfloor}(\mu_{1})\varphi_{K},\underline{\operatorname{ad}}_{A}^{\lfloor\frac{k}{2}\rfloor}(\mu_{2})\varphi_{1}\right\rangle-\\\
(-1)^{\lfloor\frac{k+1}{2}\rfloor}\left\langle\underline{\operatorname{ad}}^{\lfloor\frac{k}{2}\rfloor}_{A}(\mu_{2})\varphi_{K},\underline{\operatorname{ad}}^{\lfloor\frac{k+1}{2}\rfloor}_{A}(\mu_{1})\varphi_{1}\right\rangle.$

Once again, using the symmetry/skew-symmetry of the operator,

$\gamma_{k}=(-1)^{\lfloor\frac{k}{2}\rfloor+\lfloor\frac{k+1}{2}\rfloor}\left\langle\left[\underline{\operatorname{ad}}_{A}^{\lfloor\frac{k+1}{2}\rfloor}(\mu_{1}),\underline{\operatorname{ad}}_{A}^{\lfloor\frac{k}{2}\rfloor}(\mu_{2})\right]\varphi_{1},\varphi_{K}\right\rangle.$

This equality completes the proof. ∎

###  A.3 Some elements of proof in finite dimension

####  A.3.1 A proof of Theorem 2.14 via Sussmann’s $S(\theta)$-condition

In this section, we present a proof of Theorem 2.14 relying on Sussmann’s
$S(\theta)$-sufficient condition (see [45, Theorem 7.3]), recalled in
Proposition A.12 below (see [28, Theorem 3.29]), for which we need the
following definition:

######  Definition A.11.

The map $\sigma:\operatorname{Br}(X)\mapsto\mathcal{L}(X)$ is defined by
$\sigma(b)=\textsc{e}(b)+\pi(\textsc{e}(b))$, where
$\pi:\mathcal{L}(X)\mapsto\mathcal{L}(X)$ is the unique morphism of Lie
algebra such that $\pi(X_{0})=X_{0}$, $\pi(X_{1})=X_{2}$ and
$\pi(X_{2})=X_{1}$.

For instance, if $b=(X_{1},(X_{1},X_{0}))$ then
$\sigma(b)=[X_{1},[X_{1},X_{0}]]+[X_{2},[X_{2},X_{0}]]$.

######  Proposition A.12 (Sussmann’s $S(\theta)$-condition).

Let $f_{0},f_{1},f_{2}$ be analytic vector fields on a neighborhood of $0$ in
$\mathbb{R}^{d}$ that satisfy the Lie algebra rank condition,
$\mathcal{L}(f)(0)=\mathbb{R}^{d}$. If, there exists $\theta\in[0;1]$ such
that, for every $\mathfrak{b}\in\operatorname{Br}(X)$ with
$n_{0}(\mathfrak{b})$ odd and both $n_{1}(\mathfrak{b})$ and
$n_{2}(\mathfrak{b})$ even, we have

$f_{\sigma(\mathfrak{b})}(0)\in\text{Span}\\{f_{b}(0);b\in\operatorname{Br}(X),n(b)+\theta
n_{0}(b)<n(\mathfrak{b})+\theta n_{0}(\mathfrak{b})\\}$ (A.9)

then the system $\dot{x}=f_{0}(x)+uf_{1}(x)+vf_{2}(x)$ is $L^{\infty}$-STLC.

###### Proof of Theorem 2.14 thanks to Proposition A.12.

We prove by induction on $m\in\mathbb{N}$ that for every $d\in\mathbb{N}^{*}$,
the assumptions of Theorem 2.14 imply $W^{m,\infty}$-STLC.

Initialization for $m=0$: we prove $L^{\infty}$-STLC by applying Proposition
A.12. First (2.7) gives the Lie algebra rank condition,

$\text{Span}(f_{b_{1}}(0),\cdots f_{b_{d}}(0))=\mathbb{R}^{d},$

with $b_{1},\cdots,b_{r}\in\mathcal{B}_{1}$ such that
$\text{Span}(f_{b_{1}}(0),\cdots f_{b_{r}}(0))=S_{1}(f)(0)$ and
$b_{r+1},\cdots,b_{d}\in\mathcal{B}_{2,good}$. One considers

$0<\theta\leqslant\frac{1}{\underset{i\in\llbracket
1,d\rrbracket}{\max}|b_{i}|}.$

Let $L=\underset{i\in\llbracket r+1,d\rrbracket}{\max}|b_{i}|$. Let
$\mathfrak{b}\in\operatorname{Br}(X)$ be such that $n_{0}(\mathfrak{b})$ is
odd and both $n_{1}(\mathfrak{b})$ and $n_{2}(\mathfrak{b})$ are even. Then,

  1. 1.

Either, $n(\mathfrak{b})\geqslant 4$ and then (2.7) gives the compensation
because, for all $i\in\llbracket 1,d\rrbracket$,

$n(b_{i})+\theta n_{0}(b_{i})\leqslant 2+\theta\left(\underset{i\in\llbracket
1,d\rrbracket}{\max}|b_{i}|-1\right)<3<4+\theta
n_{0}(\mathfrak{b})=n(\mathfrak{b})+\theta n_{0}(\mathfrak{b}).$

  2. 2.

Or, $n(\mathfrak{b})=2$ i.e.
$(n_{1}(\mathfrak{b}),n_{2}(\mathfrak{b}))\in\\{(2,0),(0,2)\\}$ i.e.
$\sigma(\mathfrak{b})\in\text{Span}\\{\mathcal{B}_{2,bad}\\}$. Thus:

    1. a.

If $|\mathfrak{b}|\leqslant L$, (2.8) gives, for all $i\in\llbracket
1,r\rrbracket$,

$n(b_{i})+\theta n_{0}(b_{i})=1+\theta n_{0}(b_{i})<2\leqslant
n(\mathfrak{b})+\theta n_{0}(\mathfrak{b}).$

    2. b

If $|\mathfrak{b}|>L$, the LARC (2.7) gives once again the compensation
because,

$\begin{array}[]{rl}n(b_{i})+\theta n_{0}(b_{i})\leqslant
1+\theta\left(|b_{i}|-1\right)<2+\theta
n_{0}(\mathfrak{b})=n(\mathfrak{b})+\theta n_{0}(\mathfrak{b}),&\text{ for
}i\in\llbracket 1,r\rrbracket,\\\ n(b_{i})+\theta n_{0}(b_{i})\leqslant
2+\theta\left(L-2\right)<2+\theta(L-1)\leqslant n(\mathfrak{b})+\theta
n_{0}(\mathfrak{b}),&\text{ for }i\in\llbracket r+1,d\rrbracket.\end{array}$

Theses points prove (A.9).

Heredity. We assume the result is proved up to some $m\in\mathbb{N}$. We
consider the extended system:

$\left\\{\begin{array}[]{l}\dot{x}=f_{0}(x)+uf_{1}(x)+vf_{2}(x)\\\
\dot{u}=w\\\ \dot{v}=\omega\end{array}\right.$

with state $Y=(x,u,v)\in\mathbb{R}^{d+2}$ and control $(w,\omega)$, i.e.
$\displaystyle\frac{\mathrm{d}Y}{\mathrm{d}t}=F_{0}(Y)+wF_{1}(Y)+\omega
F_{2}(Y)$ with:

$F_{0}(x,u,v)=\begin{pmatrix}f_{0}(x)+uf_{1}(x)+vf_{2}(x)\\\ 0\\\
0\end{pmatrix},\quad F_{1}(x,u,v)=\begin{pmatrix}0\\\ 1\\\
0\end{pmatrix},\quad F_{2}(x,u,v)=\begin{pmatrix}0\\\ 0\\\ 1\end{pmatrix}.$

Then, for every $i\in\llbracket 1,2\rrbracket$,

$F_{M_{1}^{i}}(Y)=[F_{i},F_{0}](Y)=\begin{pmatrix}f_{i}(x)\\\ 0\\\
0\end{pmatrix}.$

Thus, when computing $F_{M_{j}^{i}}$ for $j\geqslant 2$, the derivatives with
respect to $u$ and $v$ never come into play, and we obtain:

$\forall j\geqslant 1\,,\quad
F_{M_{j}^{i}}(Y)=\begin{pmatrix}\underline{\operatorname{ad}}_{f_{0}+uf_{1}+vf_{2}}^{j-1}(f_{i})(x)\\\
0\\\ 0\end{pmatrix},$

where $\underline{\operatorname{ad}}$ denotes the right-bracketing:
$\underline{\operatorname{ad}}_{f}(g)=[g,f]$. Thus, for every $i\in\llbracket
1,2\rrbracket$ and $l\in\mathbb{N}$, $F_{W_{1,l}^{i}}=0$ and for every
$j\geqslant 2$,

$F_{W_{j,l}^{i}}(Y)=\begin{pmatrix}\underline{\operatorname{ad}}_{f_{0}+uf_{1}+vf_{2}}^{l}[\underline{\operatorname{ad}}_{f_{0}+uf_{1}+vf_{2}}^{j-2}(f_{i}),\underline{\operatorname{ad}}_{f_{0}+uf_{1}+vf_{2}}^{j-1}(f_{i})](x)\\\
0\\\ 0\end{pmatrix}.$

Moreover, for every $l\in\mathbb{N}$, $F_{C_{0,l}}=F_{C_{1,l}}=0$ and, for
every $j\geqslant 2$,

$F_{C_{j,l}}(Y)=\begin{pmatrix}(-1)^{j}\underline{\operatorname{ad}}_{f_{0}+uf_{1}+vf_{2}}^{l}[\underline{\operatorname{ad}}_{f_{0}+uf_{1}+vf_{2}}^{\left\lfloor\frac{j+1}{2}\right\rfloor-1}(f_{1}),\underline{\operatorname{ad}}_{f_{0}+uf_{1}+vf_{2}}^{\left\lfloor\frac{j}{2}\right\rfloor-1}(f_{2})](x)\\\
0\\\ 0\end{pmatrix}.$

In particular, for every $i\in\llbracket 1,2\rrbracket$, $j\geqslant 1$,
$F_{M_{j}^{i}}(0)=\begin{pmatrix}f_{M_{j-1}^{i}}(0)\\\ 0\\\ 0\end{pmatrix}.$
Moreover, for every $i\in\llbracket 1,2\rrbracket$, $j\geqslant 2$,
$l\in\mathbb{N}$,

$F_{W_{j,l}^{i}}(0)=\begin{pmatrix}f_{W_{j-1,l}^{i}}(0)\\\ 0\\\
0\end{pmatrix},\qquad F_{C_{j,l}}(0)=\begin{pmatrix}f_{C_{j-1,l}}(0)\\\ 0\\\
0\end{pmatrix}.$

Thus the extended system satisfies the assumptions of Theorem 2.14 with
$d+2\to d$. By the induction assumption, the extended system is
$W^{m,\infty}$-STLC. By choosing trajectories of the extended system starting
and finishing at $u=0$, one concludes that the initial system
$\dot{x}=f_{0}(x)+uf_{1}(x)+vf_{2}(x)$ is $W^{m+1,\infty}$-STLC. ∎

####  A.3.2 Proof of the representation formula of Proposition 2.19

######  Definition A.13.

We consider the (unital associative) algebra $\widehat{\mathcal{A}}(X)$ of
formal power series generated by $\mathcal{A}(X)$. An element
$a\in\widehat{\mathcal{A}}(X)$ is a sequence $a=(a_{n})_{n\in\mathbb{N}}$
written $a=\sum_{n\in\mathbb{N}}a_{n}$, where $a_{n}\in\mathcal{A}_{n}(X)$
with, in particular, $a_{0}\in\mathbb{R}$ being its constant term. We also
define the Lie algebra of formal Lie series $\widehat{\mathcal{L}}(X)$ as the
Lie algebra of formal power series $a\in\widehat{\mathcal{A}}(X)$ for which
$a_{n}\in\mathcal{L}(X$ for each $n\in\mathbb{N}$.

We consider the formal differential equation

$\left\\{\begin{aligned} \dot{y}(t)&=y(t)(X_{0}+u(t)X_{1}+v(t)X_{2}),\\\
x(0)&=1,\end{aligned}\right.$ (A.10)

whose solution is a formal series valued function
$y:\mathbb{R}_{+}\rightarrow\widehat{\mathcal{A}}(X)$ (see [9, section
$2.2$]). By [9, Section $2.4$, Theorem 41], for every $t>0$, there exists
$\mathcal{Z}_{\infty}(t,X,u,v)\in\widehat{L}(X)$ such that

$y(t)=\exp(tX_{0})\exp\left(\mathcal{Z}_{\infty}(t,X,u,v)\right).$ (A.11)

If $\mathcal{B}\subset\operatorname{Br}(X)$ is a Hall set, there exists a
unique family $(\eta_{b})_{b\in\mathcal{B}}$ of maps $\mathbb{R}_{+}\times
L^{1}_{loc}(\mathbb{R}_{+},\mathbb{R})^{2}\rightarrow\mathbb{R}$, called
coordinates of the pseudo first kind associated with $\mathcal{B}$ and they
satisfy the announced homogeneity properties. Moreover, by [9, Section 2.5],
for every $t>0$ and $u,v\in L^{1}((0,t),\mathbb{R})$,

$y(t)=\underset{b\in\mathcal{B}}{\overset{\leftarrow}{\prod}}e^{\xi_{b}(t,u,v)b}.$
(A.12)

Let $\mathcal{B}$ be a Hall set as in Proposition 2.12. We deduce from (A.11),
(A.12) and the maximality of $X_{0}$ in $\mathcal{B}$ that

$\exp\left(\mathcal{Z}_{\infty}(t,X,u,v)\right)=\underset{b\in\mathcal{B}\setminus\\{X_{0}\\}}{\overset{\leftarrow}{\prod}}e^{\xi_{b}(t,u,v)b}.$

By applying the multivariate CBHD formula [9, Proposition 34] to this
expression, one obtains $\eta_{b}=\xi_{b}$ for every $b\in\mathcal{B}_{1}$ and
for every $b\in\mathcal{B}_{2}$,

$\eta_{b}(t,u,v)=\xi_{b}(t,u,v)+\frac{1}{2}\sum\delta_{(b_{1},b_{2}),b}\xi_{b_{1}}(t,u,v)\xi_{b_{2}}(t,u,v)$

where the sum is indexed by the set
$\\{b_{1}>b_{2}\in\mathcal{B}_{1};b\in\text{supp}(b_{1},b_{2})\\}$ and
$\delta_{(b_{1},b_{2}),b}$ denotes the coefficient of $b$ in the expansion of
$(b_{1},b_{2})$ on the basis $\mathcal{B}$. In particular this sum is finite
and involves only elements $b_{1},b_{2}\in\mathcal{B}_{1}$ such that
$n_{0}(b_{1})+n_{0}(b_{1})=n_{0}(b)$.

The estimate (2.15) is proved in [9, Proposition 161]. The absolute
convergence is proved in [9, Proposition 103] and relies on [8, Theorem 1.9].

###  A.4 Some lemmas about ODEs

######  Lemma A.14.

Let $\delta>0$ and $z\in\mathcal{C}^{1}(B(0,{\delta});\mathbb{R}^{d})$ be such
that $\left\|z\right\|_{\infty}\leqslant\delta$. We note $x(\cdot,z,0)$ the
solution to the system: $\left\\{\begin{matrix}x^{\prime}&=&z(x)\\\
x(0)&=&0\end{matrix}\right.$. Then,

$\left\|x(1;z,0)-z(0)\right\|\leqslant\left\|z(0)\right\|\left\|Dz\right\|_{\infty}e^{\left\|Dz\right\|_{\infty}}.$

###### Proof.

This lemma is proved in [9, Lemma 160]. ∎

######  Lemma A.15.

Let $\delta>0$, and $f_{0},f_{1},f_{2}:B(0,2\delta)\to\mathbb{R}^{d}$ be real-
analytic functions, with $f_{0}(0)=0$. There exists $r(\delta)>0$, such that,
for all $T_{1}<T<T_{1}+1$, $p\in B\left(0,r(\delta)\right)$, and $u,v\in
L^{\infty}(T_{1},T)$ with
$\left\|u\right\|_{L^{\infty}},\left\|v\right\|_{L^{\infty}}<r(\delta)$, one
has:

$\forall t\in[T_{1},T],\quad x(T;(u,v),p,T_{1})\in B(0,\delta).$ (A.13)

###### Proof.

This result is linked to continuity with respect to the initial condition ∎

######  Corollary A.16.

Under the same assumptions as the previous lemma, there exists $r(\delta)>0$,
$C(\delta)>0$, such that, for all $T_{1}<T<T_{1}+1$, $p\in
B\left(0,r(\delta)\right)$, and $u,v\in L^{\infty}(T_{1},T)$ with
$\left\|u\right\|_{L^{\infty}},\left\|v\right\|_{L^{\infty}}<r(\delta)$, one
has:

$\left\|x(T;(u,v),p,T_{1})\right\|\leqslant
C(\delta)\left(\left\|p\right\|+\left\|(u,v)\right\|_{L^{\infty}}\right).$

###### Proof.

This statement is based on the previous Lemma and the Grönwall inequality. ∎

######  Lemma A.17.

Let $f_{0}:B(0,2\delta)\to\mathbb{R}^{d}$, be a real-analytic function, for
$\delta>0$, with $f_{0}(0)=0$. There exists $C>0$, $\varepsilon>0$ such that,
for all $T_{1}<T<T_{1}+1$ and $p\in B\left(0,\varepsilon\right)$,

$\left\|x(T;(0,0),p,T_{1})-e^{(T-T_{1})Df_{0}(0)}p\right\|\leqslant
C\left\|p\right\|^{2}.$

###### Proof.

By definition, there exists $h>0$, s.t. for all $x\in B(0,h)$,

$\left\|f_{0}(x)-f_{0}(0)-Df_{0}(0)x\right\|\leqslant\left(\frac{1}{2}\left\|D^{2}f_{0}(0)\right\|+1\right)\left\|x\right\|^{2}.$
(A.14)

Let $p\in B(0,\varepsilon)$ with
$\varepsilon=\min\left(r(\delta),r(h)\right)$. Using the Duhamel formula,

$\left\|x(T;(0,0),p,T_{1})-e^{(T-T_{1})Df_{0}(0)}p\right\|\leqslant\\\
\int_{T_{1}}^{T}\left\|e^{(t-s)Df_{0}(0)}\right\|\left\|f_{0}(x(s,(0,0),p,T_{1}))-Df_{0}(0)x(s,(0,0),p,T_{1})\right\|\mathrm{d}s.$

Thank to Lemma (A.15) one can use the inequality (A.14), and Lemma (A.16)
gives the result. ∎

######  Lemma A.18.

Let $\delta>0$, and $f_{0},f_{1},f_{2}:B(0,2\delta)\to\mathbb{R}^{d}$ be real-
analytic functions, with $f_{0}(0)=0$. There exists $C,\varepsilon>0$, such
that, for all $T_{1}<T<T_{1}+1$, $p\in B\left(0,\varepsilon\right)$, and
$u,v\in L^{\infty}(T_{1},T)$ with
$\left\|u\right\|_{L^{\infty}},\left\|v\right\|_{L^{\infty}}<\varepsilon$, one
has:

$\left\|x(T;(u,v),p,T_{1})-x(T;(u,v),0,T_{1})-x(T;(0,0),p,T_{1})\right\|\leqslant
C\left(\left\|p\right\|\left\|(u,v)\right\|_{L^{\infty}}+\left\|p\right\|^{2}\right).$

###### Proof.

Let $p\in B\left(0,\varepsilon\right)$, $u,v\in L^{\infty}(T_{1},T)$, small
enough. For simplicity, we use the notations: $x:=x(\cdot;(u,v),p,T_{1})$,
$x_{u,v}:=x(\cdot;(u,v),0,T_{1})$, and $x_{p}:=x(\cdot;(0,0),p,T_{1})$. Let
$z:=x-x_{u,v}-x_{p}$. By definition, $z$ is solution to the following control-
affine system:

$z^{\prime}=\left(f_{0}(z+x_{u,v}+x_{p})-f_{0}(x_{u,v}+x_{p})\right)+\left(f_{0}(x_{u,v}+x_{p})-f_{0}(x_{u,v})-f_{0}(x_{p})\right)+\\\
u\left(f_{1}(z+x_{u,v}+x_{p})-f_{1}(x_{u,v})\right)+v\left(f_{2}(z+x_{u,v}+x_{p})-f_{2}(x_{u,v})\right),$

with $z(T_{1})=0$. We estimate each term of this formulation:

  1. 1.

The first one scales like $\left\|z\right\|$.

  2. 2.

The second one is shaped like
$\left\|p\right\|\left(\left\|p\right\|+\left\|(u,v)\right\|_{\infty}\right)$.

  3. 3.

The other terms are like
$\left\|(u,v)\right\|_{\infty}\left(\left\|z\right\|+\left\|p\right\|\right)$.

The Grönwall lemma gives the result. ∎

## Acknowledgement

The author would like to take particular care in thanking Karine Beauchard and
Frédéric Marbach for the many discussions that brought this article to life.

The author acknowledges support from grants ANR-20-CE40-0009 (Project TRECOS)
and ANR-11-LABX-0020 (Labex Lebesgue), as well as from the Fondation Simone et
Cino Del Duca – Institut de France.

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