Robustness under control sampling of reachability in fixed time for nonlinear control systems

Abstract

Under a regularity assumption we prove that reachability in fixed time for nonlinear control systems is robust under control sampling.

Appendices

An example

We develop here an example inspired from [17, Section II], showing that the converse of the geometric Pontryagin maximum principle is not true in general and that, given a control $u \in {\mathcal {U}}\cap \mathrm {L}^\infty ([0,T],\mathrm {U})$, the condition that $x_u(T)$ belongs to the interior of the $\mathrm {L}^\infty _\mathrm {U}$-accessible set is not a sufficient condition for Property (${\mathcal {R}}^{\mathrm{{unif}}}_u$), even if $\mathrm {U}$ is convex.

Take $T = n = m = 2$ and $\mathrm {U}= {\mathbb {R}}^2$. Take $g_1 \in \mathrm {C}([0,2],{\mathbb {R}})$ be a continuous function that is positive on the interval [0, 1) and vanishing on the interval [1, 2]. Take $g_2 \in \mathrm {L}^\infty ([0,2],{\mathbb {R}})$ be arbitrarily fixed and $g_3 \in \mathrm {AC}([0,2],{\mathbb {R}})$ be defined by $g_3(t) = \int _0^t g_1(\xi )g_2(\xi ) \, \mathrm{d}\xi$ for all $t \in [0,2]$. Note that $g_3$ is constant on the interval [1, 2]. We denote by G the corresponding constant values. We set $x^0 = 0_{{\mathbb {R}}^2}$ and

$$\begin{aligned} f((x_1,x_2),(u_1,u_2),t) = \begin{pmatrix} g_1(t)u_1 + g_1(2-t) \Big ( (x_1-G)^2 + x_2^2 \Big ) u_1 \\ \Big (x_1 - g_3(t) \Big )^2 + g_1(2-t) \Big ( (x_1-G)^2 + x_2^2 \Big ) u_2 \end{pmatrix} \end{aligned}$$

for all $((x_1,x_2),(u_1,u_2),t) \in {\mathbb {R}}^2 \times {\mathbb {R}}^2 \times [0,2]$.

Claim 1

The point (G, 0) is an equilibrium of the control system on the interval [1, 2], independently of the control.

Proof

Since $g_1(t) = 0$ and $g_3(t) = G$ for all $t \in [1,2]$, we have $f((G,0),u,t) = 0_{{\mathbb {R}}^2}$ for all $(u,t) \in {\mathbb {R}}^2 \times [1,2]$. $\square$

Claim 2

Let $u \in \mathrm {L}^\infty ([0,2],{\mathbb {R}}^2)$ satisfying $u_1(t)=g_2(t)$ for a.e. $t \in [0,1]$. Then $u \in {\mathcal {U}}$ and $x_u = (g_3,0)$. In particular $x_u(2) = (G,0)$.

Proof

Since $g_1(2-\xi ) = 0$ for all $\xi \in [0,1]$, $x_{u,1}(t) = \int _0^t g_1(\xi )u_1(\xi )\, \mathrm{d}\xi = \int _0^t g_1(\xi )g_2(\xi )\, \mathrm{d}\xi = g_3(t)$ for all $t \in [0,1]$. From the second coordinate, we obtain that $x_{u,2}(t) = 0$ for all $t \in [0,1]$. Since $x_u(1) = (g_3(1),0) = (G,0)$, we get from Claim 1 that $x_u(t) = (G,0) = (g_3(t),0)$ for all $t \in [1,2]$. $\square$

Claim 3

Let $u \in {\mathcal {U}}$ such that $x_u(T') = (G,0)$ for some $T' \in [1,2]$. Then $u_1(t) = g_2(t)$ for a.e. $t \in [0,1]$.

Proof

By Claim 1, $x_u(1) = (G,0)$. Since $g_1(2-\xi ) = 0$ for all $\xi \in [0,1]$, we get that $0 = x_{u,2}(1) = \int _0^1 (x_{u,1}(\xi ) - g_3(\xi ) )^2 \, \mathrm{d}\xi$ and thus $x_{u,1}(t) = g_3(t)$ for all $t \in [0,1]$. Deriving this equality leads to $g_1(t)u_1(t) = g_1(t)g_2(t)$ for a.e. $t \in [0,1]$. Since $g_1$ is positive on the interval [0, 1), we get that $u_1(t) = g_2(t)$ for a.e. $t \in [0,1]$. $\square$

Claim 4

The end-point mapping is surjective.

Proof

Let $x^1 \in {\mathbb {R}}^2$. Let us prove that there exists $u \in {\mathcal {U}}$ such that $\mathrm {E}(u) = x_u(2) = x^1$. If $x^1 = (G,0)$, from Claim 2, it is sufficient to take any control $u \in \mathrm {L}^\infty ([0,2],{\mathbb {R}}^2)$ which satisfies $u_1(t) = g_2(t)$ for a.e. $t \in [0,1]$. In the rest of this proof, we focus on the case $x^1 \ne (G,0)$.

Consider a function $g_4 \in \mathrm {L}^\infty ([0,\frac{3}{2}],{\mathbb {R}})$ such that the measure of $\{ t \in [0,1] \mid g_4(t) \ne g_2 (t) \}$ is positive and such that the $\mathrm {L}^\infty$-norm of $g_4 - g_2$ on [0, 1] is small enough to guarantee that any control $u \in \mathrm {L}^\infty ([0,2],{\mathbb {R}}^2)$ which satisfies $u_1(t) = g_4(t)$ for a.e. $t \in [0,\frac{3}{2}]$ is admissible, i.e., $u \in {\mathcal {U}}$. This is possible by Claim 2, since ${\mathcal {U}}$ is an open subset of $\mathrm {L}^\infty ([0,2],{\mathbb {R}}^2)$. Take such a control u (which is only determined on the interval $[0,\frac{3}{2}]$ at this step). By Claim 3, $x_u( \frac{3}{2} ) \ne (G,0)$. Consider now a $\mathrm {C}^1$ function $\varrho : [\frac{3}{2},2] \rightarrow {\mathbb {R}}^2$ which satisfies $\varrho (\frac{3}{2}) = x_u( \frac{3}{2} )$, $\varrho (2) = x^1$ and $\varrho (t) \ne (G,0)$ for all $t \in [\frac{3}{2},2]$. We determine the control u on $[0,\frac{3}{2}]$ as

$$\begin{aligned} u_1 (t) = \dfrac{\dot{\varrho _1}(t)}{\overline{\varrho }(t)}, \quad u_2 (t) = \dfrac{\dot{\varrho _2}(t) - ( \varrho _1(t) - g_3(t) )^2 }{\overline{\varrho }(t)}, \end{aligned}$$

where $\overline{\varrho }(t)=g_1(2-t) ( (\varrho _1(t) - G)^2 + \varrho _2(t)^2 )$ for a.e. $t \in [\frac{3}{2},2]$. The control u belongs to $\mathrm {L}^\infty ([0,2],{\mathbb {R}}^2)$ and $x_u = \varrho$ along $[\frac{3}{2},2]$. Thus $\mathrm {E}(u) = x_u(2) = \varrho (2) = x^1$. $\square$

Let us prove that the converse of the geometric Pontryagin maximum principle is not true in general. Take a control $u \in \mathrm {L}^\infty ([0,T],{\mathbb {R}}^2)$ which satisfies $u_1(t)=g_2(t)$ for a.e. $t \in [0,1]$. By Claims 2 and 4, we have $u \in {\mathcal {U}}$ and $x_u(2)$ belongs to the interior of the $\mathrm {L}^\infty _{{\mathbb {R}}^2}$-accessible set. Consider the constant function $p : [0,2] \rightarrow {\mathbb {R}}^2$ defined by $p(t)=(0,1) \ne 0_{{\mathbb {R}}^2}$ for all $t \in [0,2]$. One can easily check that $(x_u,u,p)$ is a nontrivial strong ${\mathbb {R}}^2$-extremal lift of $(x_u,u)$ and thus u is strongly ${\mathbb {R}}^2$-singular by Proposition 2.6.

We now prove that, given a control $u \in {\mathcal {U}}\cap \mathrm {L}^\infty ([0,T],\mathrm {U})$, the condition that $x_u(T)$ belongs to the interior of the $\mathrm {L}^\infty _\mathrm {U}$-accessible set is not a sufficient condition for Property (${\mathcal {R}}^{\mathrm{{unif}}}_u$), even if $\mathrm {U}$ is convex. Take $g_2(t)=t$ for a.e. $t \in [0,1]$ (which is not piecewise constant). Even if (G, 0) belongs to the interior of the $\mathrm {L}^\infty _{\mathrm {U}}$-accessible set (Claim 4), we easily infer from Claim 3 that (G, 0) is not $\mathrm {PC}^{\mathbb {T}}_{\mathrm {U}}$-reachable in time T from $x^0$ for any partition ${\mathbb {T}}$ of [0, T]. Hence Property (${\mathcal {R}}_u$) is not satisfied, and neither is the stronger Property (${\mathcal {R}}^{\mathrm{{unif}}}_u$).

A general result on $\mathrm {L}^s$-approximation by piecewise constant functions

Proposition B.1

Let $1 \le s < +\infty$. Given any $u \in \mathrm {L}^\infty ([0,T],\mathrm {U})$ and any $\varepsilon > 0$, there exists a threshold $\delta > 0$ such that, for any partition ${\mathbb {T}}$ of [0, T] satisfying $\Vert {\mathbb {T}}\Vert \le \delta$, there exists $v \in \mathrm {PC}^{\mathbb {T}}([0,T],\mathrm {U})$ such that $\Vert v - u \Vert _{\mathrm {L}^s} \le \varepsilon$ and $\Vert v \Vert _{\mathrm {L}^\infty } \le \Vert u \Vert _{\mathrm {L}^\infty }$.

Proof

Let $u \in \mathrm {L}^\infty ([0,T],\mathrm {U})$ and $\varepsilon > 0$. By the Lusin theorem [23], there exists a compact subset $\mathrm {K}_\varepsilon \subset [0,T]$ such that $( 2 \Vert u \Vert _{\mathrm {L}^\infty } )^s \mu ( [0,T] \backslash \mathrm {K}_\varepsilon ) \le \varepsilon ^s/2$, where $\mu$ is the Lebesgue measure, and such that u is continuous on $\mathrm {K}_\varepsilon$. By uniform continuity of u on $\mathrm {K}_\varepsilon$, there exists $\delta > 0$ such that $\Vert u(\xi _2) - u(\xi _1) \Vert _{{\mathbb {R}}^m} \le \frac{\varepsilon }{(2T)^{1/s}}$ for all $\xi _1$, $\xi _2 \in \mathrm {K}_\varepsilon$ satisfying $\vert \xi _2 - \xi _1 \vert \le \delta$. Now, let ${\mathbb {T}}= \{ t_i \}_{i=0,\ldots ,N}$ be a partition of [0, T] such that $\Vert {\mathbb {T}}\Vert \le \delta$. We set

$$\begin{aligned} I = \{ i \in \{ 0,\ldots ,N-1 \} \mid \mu (\mathrm {K}_\varepsilon \cap [t_i,t_{i+1}) > 0 \}. \end{aligned}$$

For every $i \in I$, we consider some $\xi _i \in \mathrm {K}_\varepsilon \cap [t_i,t_{i+1})$ such that $u(\xi _i) \in \mathrm {U}$ and $\Vert u(\xi _i) \Vert _{{\mathbb {R}}^m} \le \Vert u \Vert _{\mathrm {L}^\infty }$. We also consider some $\omega \in \mathrm {U}$ such that $\Vert \omega \Vert _{{\mathbb {R}}^m} \le \Vert u \Vert _{\mathrm {L}^\infty }$. We now define

$$\begin{aligned} v(t) = \left\{ \begin{array}{lcl} u(\xi _i) &{} \text {if} &{} t \in [t_i,t_{i+1}) \text { with } i \in I, \\ \omega &{} \text {if} &{} t \in [t_i,t_{i+1}) \text { with } i \notin I, \end{array} \right. \end{aligned}$$

for every $t \in [0,T]$. In particular we have $v \in \mathrm {PC}^{\mathbb {T}}([0,T],\mathrm {U})$ and $\Vert v \Vert _{\mathrm {L}^\infty } \le \Vert u \Vert _{\mathrm {L}^\infty }$. Finally we get that

$$\begin{aligned} \Vert v - u \Vert ^s_{\mathrm {L}^s}= & {} \int _{[0,T] \backslash \mathrm {K}_\varepsilon } \Vert v(t) - u(t) \Vert _{{\mathbb {R}}^m}^s \; \mathrm{d}t + \displaystyle \sum _{i=0}^{N-1} \int _{\mathrm {K}_\varepsilon \cap [t_i,t_{i+1})} \Vert v(t) - u(t) \Vert _{{\mathbb {R}}^m}^s \; \mathrm{d}t \\\le & {} (2 \Vert u \Vert _{\mathrm {L}^\infty })^s \mu ( [0,T] \backslash \mathrm {K}_\varepsilon ) + \displaystyle \sum _{i \in I} \int _{\mathrm {K}_\varepsilon \cap [t_i,t_{i+1})} \Vert u(\xi _i) - u(t) \Vert _{{\mathbb {R}}^m}^s \; \mathrm{d}t \\\le & {} \dfrac{\varepsilon ^s}{2} + \dfrac{\varepsilon ^s}{2T} \displaystyle \sum _{i \in I} \mu ( \mathrm {K}_\varepsilon \cap [t_i,t_{i+1}) ) \le \varepsilon ^s , \end{aligned}$$

which concludes the proof. $\square$

Note that Proposition B.1 is not true with $s=+\infty$, as shown in the following Fuller-type example [11].

Example B.1

Take $T = 1$, $m=1$ and $\mathrm {U}= {\mathbb {R}}$. Consider the oscillating function $u \in \mathrm {L}^\infty ([0,T],\mathrm {U})$ defined by $u(t)=1$ for a.e. $t \in (\frac{1}{k+1},\frac{1}{k}]$ for all even $k \in {\mathbb {N}}\backslash \{0\}$ and $u(t)=0$ for a.e. $t \in (\frac{1}{k+1},\frac{1}{k}]$ for all odd $k \in {\mathbb {N}}\backslash \{0\}$. We have $\Vert v - u \Vert _{\mathrm {L}^\infty } \ge \frac{1}{2}$ for all $v \in \mathrm {PC}^{\mathbb {T}}([0,T],\mathrm {U})$ and all partitions ${\mathbb {T}}$ of [0, T].

Corollary B.1

Let $1 \le s < +\infty$. Given any $u \in \mathrm {L}^s([0,T],\mathrm {U})$ and any $\varepsilon > 0$, there exists a threshold $\delta > 0$ such that, for any partition ${\mathbb {T}}$ of [0, T] satisfying $\Vert {\mathbb {T}}\Vert \le \delta$, there exists $v \in \mathrm {PC}^{\mathbb {T}}([0,T],\mathrm {U})$ such that $\Vert v - u \Vert _{\mathrm {L}^s} \le \varepsilon$.

Proof

Let $u \in \mathrm {L}^s([0,T],\mathrm {U})$ and $\varepsilon > 0$. We fix some $\omega \in \mathrm {U}$ and we define $C_k = \{ t \in [0,T] \mid \Vert u(t) \Vert _{{\mathbb {R}}^m} \ge k \}$ and

$$\begin{aligned} u_k (t) = \left\{ \begin{array}{lcl} u(t) &{} \text {if} &{} t \notin C_k, \\ \omega &{} \text {if} &{} t \in C_k, \end{array} \right. \end{aligned}$$

for a.e. $t \in [0,T]$ and for every $k \in {\mathbb {N}}$. In particular $u_k \in \mathrm {L}^\infty ([0,T],\mathrm {U})$ for every $k \in {\mathbb {N}}$. It is clear that $( u_k(t) - u(t) )_{k \in {\mathbb {N}}}$ converges to $0_{{\mathbb {R}}^m}$ as $k \rightarrow +\infty$ and that $\Vert u_k(t) - u(t) \Vert _{{\mathbb {R}}^m} \le \Vert \omega \Vert _{{\mathbb {R}}^m} + \Vert u(t) \Vert _{{\mathbb {R}}^m}$ for a.e. $t \in [0,T]$. By the Lebesgue dominated convergence theorem, we get that $\Vert u_k - u \Vert _{\mathrm {L}^s} \rightarrow 0$ as $k \rightarrow +\infty$. Hence, there exists $k \in {\mathbb {N}}$ such that $\Vert u_k - u \Vert _{\mathrm {L}^s} \le \frac{\varepsilon }{2}$. By Proposition B.1, there exists $\delta > 0$ such that, for any partition ${\mathbb {T}}$ of [0, T] satisfying $\Vert {\mathbb {T}}\Vert \le \delta$, there exists $v \in \mathrm {PC}^{\mathbb {T}}([0,T],\mathrm {U})$ such that $\Vert v - u_k \Vert _{\mathrm {L}^s} \le \frac{\varepsilon }{2}$ and thus $\Vert v - u \Vert _{\mathrm {L}^s} \le \Vert v - u_k \Vert _{\mathrm {L}^s} + \Vert u_k - u \Vert _{\mathrm {L}^s} \le \varepsilon$. $\square$

1.1 Averaging operators

For any partition ${\mathbb {T}}= \{ t_i \}_{i=0,\ldots ,N}$ of [0, T], we define the averaging operator ${\mathcal {I}}^{\mathbb {T}}: \mathrm {L}^1([0,T],{\mathbb {R}}^m) \rightarrow \mathrm {PC}^{\mathbb {T}}([0,T],{\mathbb {R}}^m)$ by

$$\begin{aligned} {\mathcal {I}}^{\mathbb {T}}(u)(t) = \dfrac{1}{t_{i+1} -t_i} \int _{t_i}^{t_{i+1}} u(\xi ) \, \mathrm{d}\xi \end{aligned}$$

(7)

for every $t \in [t_i,t_{i+1})$, every $i \in \{ 0,\ldots ,N-1 \}$ and every $u \in \mathrm {L}^1([0,T],{\mathbb {R}}^m)$. The aim of this section is to establish several useful properties of the averaging operators.

Let ${\mathbb {T}}= \{ t_i \}_{i=0,\ldots ,N}$ be a partition of [0, T]. The averaging operator ${\mathcal {I}}^{\mathbb {T}}$ is linear and projects any integrable function onto a piecewise constant function respecting the partition ${\mathbb {T}}$ (by averaging its value on each sampling interval $[t_i,t_{i+1})$).

Lemma B.1

Let $1 \le s \le +\infty$. For any partition ${\mathbb {T}}$ of [0, T], we have $\Vert {\mathcal {I}}^{\mathbb {T}}(u) \Vert _{\mathrm {L}^s} \le \Vert u \Vert _{\mathrm {L}^s}$ for all $u \in \mathrm {L}^s([0,T],{\mathbb {R}}^m)$.

Proof

Let ${\mathbb {T}}= \{ t_i \}_{i=0,\ldots ,N}$ be a partition of [0, T] and let $u \in \mathrm {L}^s([0,T],{\mathbb {R}}^m)$. When $s=+\infty$, the inequality $\Vert {\mathcal {I}}^{\mathbb {T}}(u) \Vert _{\mathrm {L}^\infty } \le \Vert u \Vert _{\mathrm {L}^\infty }$ is trivial. When $1 \le s < +\infty$, we get from the Hölder inequality that

$$\begin{aligned} \left\| \dfrac{1}{t_{i+1} -t_i} \int _{t_i}^{t_{i+1}} u(\xi ) \, \mathrm{d}\xi \right\| _{{\mathbb {R}}^m}^s \le \dfrac{1}{t_{i+1} -t_i} \int _{t_i}^{t_{i+1}} \Vert u(\xi ) \Vert ^s_{{\mathbb {R}}^m} \, \mathrm{d}\xi \end{aligned}$$

for all $i \in \{ 0,\ldots ,N-1 \}$, and thus $\Vert {\mathcal {I}}^{\mathbb {T}}(u) \Vert ^s_{\mathrm {L}^s} \le \Vert u \Vert ^s_{\mathrm {L}^s}$. $\square$

The next lemma is instrumental in order to approximate with a $\mathrm {L}^s$-norm (with any $1 \le s < +\infty$) any control $u \in \mathrm {L}^\infty ([0,T],{\mathbb {R}}^m)$ with piecewise constant controls.

Lemma B.2

Let $1 \le s < +\infty$. Given any $u \in \mathrm {L}^s([0,T],{\mathbb {R}}^m)$, we have $\Vert {\mathcal {I}}^{\mathbb {T}}(u) - u \Vert _{\mathrm {L}^s} \rightarrow 0$ as $\Vert {\mathbb {T}}\Vert \rightarrow 0$.

Proof

Let $u \in \mathrm {L}^s([0,T],{\mathbb {R}}^m)$ and let us prove that $\Vert {\mathcal {I}}^{\mathbb {T}}(u) - u \Vert _{\mathrm {L}^s} \rightarrow 0$ as $\Vert {\mathbb {T}}\Vert \rightarrow 0$. Our proof is based on a density argument. Therefore, in a first place, we deal with the case where u is essentially bounded, that is, $u \in \mathrm {L}^\infty ([0,T],{\mathbb {R}}^m)$. From Lemma B.1, the domination $\Vert {\mathcal {I}}^{\mathbb {T}}(u)(t) \Vert _{{\mathbb {R}}^m} \le \Vert u \Vert _{\mathrm {L}^\infty }$ is true for a.e. $t \in [0,T]$ and thus, thanks to the Lebesgue dominated convergence theorem, we only need to prove that ${\mathcal {I}}^{\mathbb {T}}(u)(\tau ) \rightarrow u(\tau )$ as $\Vert {\mathbb {T}}\Vert \rightarrow 0$ for a.e. $\tau \in [0,T]$. For this purpose we set $r(t) = \int _0^t u(\xi ) \, \mathrm{d}\xi$ for every $t \in [0,T]$ and let $\tau \in [0,T)$ being a Lebesgue point such that r is derivable at $\tau$ with ${\dot{r}}(\tau ) = u(\tau )$. Given any $\varepsilon > 0$, there exists $\delta > 0$ such that

$$\begin{aligned} \left\| \dfrac{r(t)-r(\tau )}{t - \tau } - u(\tau ) \right\| _{{\mathbb {R}}^m} \le \dfrac{\varepsilon }{2} \end{aligned}$$

for every $t \in [\tau -\delta ,\tau +\delta ] \cap [0,T] \backslash \{ \tau \}$. Take ${\mathbb {T}}$ a partition of [0, T] such that $\Vert {\mathbb {T}}\Vert \le \delta$. There exists $i \in \{ 0, \ldots , N-1 \}$ such that $\tau \in [t_i,t_{i+1})$. Then

$$\begin{aligned} \Vert {\mathcal {I}}^{\mathbb {T}}(u) (\tau ) - u(\tau ) \Vert _{{\mathbb {R}}^m}= & {} \left\| \dfrac{r(t_{i+1})-r(t_i)}{t_{i+1}-t_i} - u(\tau ) \right\| _{{\mathbb {R}}^m} \\\le & {} \left\| \dfrac{r(t_{i+1})-r(\tau )}{t_{i+1}-\tau } - u(\tau ) \right\| _{{\mathbb {R}}^m} \left| \dfrac{t_{i+1}-\tau }{t_{i+1}-t_i} \right| \\&+ \left\| \dfrac{r(\tau )-r(t_i)}{\tau -t_i} - u(\tau ) \right\| _{{\mathbb {R}}^m} \left| \dfrac{\tau -t_i}{t_{i+1}-t_i} \right| \le \varepsilon , \end{aligned}$$

which concludes the proof in the case where u is essentially bounded. Now let us deal with the general case. Take $\varepsilon > 0$ be arbitrary. By a density argument, there exists $u' \in \mathrm {L}^\infty ([0,T],{\mathbb {R}}^m)$ such that $\Vert u - u' \Vert _{\mathrm {L}^s} \le \frac{\varepsilon }{2}$. From the first step, there exists $\delta > 0$ such that $\Vert {\mathcal {I}}^{\mathbb {T}}(u') - u' \Vert _{\mathrm {L}^s} \le \frac{\varepsilon }{2}$ for any partition ${\mathbb {T}}$ of [0, T] such that $\Vert {\mathbb {T}}\Vert \le \delta$. By linearity of ${\mathcal {I}}^{\mathbb {T}}$ and by Lemma B.1, we obtain that $\Vert {\mathcal {I}}^{\mathbb {T}}(u) - u \Vert _{\mathrm {L}^s} \le \Vert {\mathcal {I}}^{\mathbb {T}}(u) - {\mathcal {I}}^T(u') \Vert _{\mathrm {L}^s} + \Vert {\mathcal {I}}^{\mathbb {T}}(u') - u' \Vert _{\mathrm {L}^s} \le \varepsilon$ for any partition ${\mathbb {T}}$ of [0, T] such that $\Vert {\mathbb {T}}\Vert \le \delta$. The proof is complete. $\square$

Our objective now is to prove that, when $\mathrm {U}$ is convex, the averaging operators project any integrable function with values in $\mathrm {U}$ onto a piecewise constant function with values in $\mathrm {U}$.

Lemma B.3

Assume that $\mathrm {U}$ is convex. If $u \in \mathrm {L}^1([0,1],\mathrm {U})$, then $\int _0^1 u(\xi ) \mathrm{d}\xi \in \mathrm {U}$.

Proof

Let $u \in \mathrm {L}^1([0,1],\mathrm {U})$ and let us prove that $\widetilde{u} \in \mathrm {U}$ where $\widetilde{u}$ is defined by $\widetilde{u} = \int _0^1 u(\xi ) \mathrm{d}\xi$. We first give a simpler argument when $\mathrm {U}$ is furthermore assumed to be closed. In that context, by the Hilbert projection theorem, we have

$$\begin{aligned} \langle \widetilde{u} - \mathrm {proj}_\mathrm {U}( \widetilde{u} ), u(\xi ) - \mathrm {proj}_\mathrm {U}(\widetilde{u} ) \rangle _{{\mathbb {R}}^m} \le 0 \end{aligned}$$

for a.e. $\xi \in [0,1]$, where $\mathrm {proj}_\mathrm {U}( \widetilde{u} ) \in \mathrm {U}$ is the projection of $\widetilde{u}$ onto $\mathrm {U}$. Integrating the above inequality over [0, 1] yields $\Vert \widetilde{u} - \mathrm {proj}_\mathrm {U}( \widetilde{u} ) \Vert _{{\mathbb {R}}^m}^2 \le 0$ and thus $\widetilde{u} = \mathrm {proj}_\mathrm {U}( \widetilde{u} ) \in \mathrm {U}$.

Now we remove the closedness assumption made on $\mathrm {U}$. Let us prove that $\widetilde{u} \in \mathrm {U}$ by strong induction on the dimension $d \in {\mathbb {N}}$ of the nonempty convex set $\mathrm {U}$. If $d=0$, the set $\mathrm {U}$ is reduced to a singleton and the result is trivial. Now consider that $d \ge 1$ and assume that the result is true at all steps from 0 to $d-1$. By contradiction assume that $\widetilde{u} \notin \mathrm {U}$. By separation, there exists $\psi \in {\mathbb {R}}^m \backslash \{ 0_{{\mathbb {R}}^m} \}$ such that $\langle \psi , \omega - \widetilde{u} \rangle _{{\mathbb {R}}^m} \le 0$ for all $\omega \in \mathrm {U}$. We infer that the null integral $\int _0^1 \langle \psi , u(\xi ) - \widetilde{u} \rangle _{{\mathbb {R}}^m} \mathrm{d}\xi$ has a nonpositive integrand. Thus this integrand is zero almost everywhere on [0, 1]. Therefore, u is with values in the convex set $\mathrm {U}\cap ( \widetilde{u} + \psi ^\bot )$, where $\psi ^\bot$ stands for the standard hyperplane defined by orthogonality with the nonzero vector $\psi$. Since $\mathrm {U}\cap ( \widetilde{u} + \psi ^\bot )$ is a nonempty convex set of dimension strictly inferior than d, thanks to our induction hypothesis we get that $\widetilde{u} \in \mathrm {U}\cap ( \widetilde{u} + \psi ^\bot )$, which raises a contradiction. $\square$

From Lemma B.3 and applying a simple affine change of variable in (7), we obtain the next proposition.

Proposition B.2

Assume that $\mathrm {U}$ is convex. If $u \in \mathrm {L}^1([0,T],\mathrm {U})$, then ${\mathcal {I}}^{\mathbb {T}}(u) \in \mathrm {PC}^{\mathbb {T}}([0,T],\mathrm {U})$ for any partition ${\mathbb {T}}$ of [0, T].

1.2 Truncated end-point mapping and $\mathrm {L}^s$-differential

For every $M > 0$, we fix a mapping $\Lambda ^M : {\mathbb {R}}^n \times {\mathbb {R}}^m \rightarrow {\mathbb {R}}$ of class $\mathrm {C}^1$ satisfying

$$\begin{aligned} \Lambda ^M(x,u) = \left\{ \begin{array}{l} 1 \text { if } (x,u) \in \overline{\mathrm {B}}_{{\mathbb {R}}^n}(0,2M) \times \overline{\mathrm {B}}_{{\mathbb {R}}^m}(0,2M), \\ 0 \text { if } (x,u) \notin \mathrm {B}_{{\mathbb {R}}^n}(0,3M) \times \mathrm {B}_{{\mathbb {R}}^m}(0,3M) . \end{array} \right. \end{aligned}$$

Let $M > 0$. When replacing the dynamics f in the control system (CS) by the truncated dynamics $f^M$, defined by $f^M (x,u,t) = \Lambda ^M(x,u) f(x,u,t)$ for all $(x,u,t) \in {\mathbb {R}}^n \times {\mathbb {R}}^m \times [0,T]$ we obtain a new control system that we denote by ($\mathrm {CS}^M$). The main difference is that, for any control $u \in \mathrm {L}^1([0,T],{\mathbb {R}}^m)$ (even unbounded), there exists a trajectory $x \in \mathrm {AC}([0,T], {\mathbb {R}}^n)$, starting at $x(0)=x^0$, such that ${\dot{x}}(t) = f^M ( x(t),u(t),t )$ for a.e. $t \in [0,T]$. In that case the trajectory x is unique and will be denoted by $x^M_u$. We now introduce, for any $1 \le s \le +\infty$, the truncated end-point mapping $\mathrm {E}^M : \mathrm {L}^s([0,T],{\mathbb {R}}^m)\rightarrow {\mathbb {R}}^n$ defined by $\mathrm {E}^M(u) = x^M_u(T)$ for all $u \in \mathrm {L}^s([0,T],{\mathbb {R}}^m)$. Note that the next proposition, derived from standard techniques in ordinary differential equations theory, is true for any $1 < s \le +\infty$. The case $s=1$ is discussed in Remark B.2.

Proposition B.3

Let $1 < s \le +\infty$ and $M > 0$. The truncated end-point mapping $\mathrm {E}^M:\mathrm {L}^s([0,T],{\mathbb {R}}^m)\rightarrow {\mathbb {R}}^n$ is of class $\mathrm {C}^1$ and its Fréchet differential is given by

$$\begin{aligned} \mathrm {D}\mathrm {E}^M (u) \cdot v = w^{u,M}_v(T) \end{aligned}$$

(8)

for all u, $v \in \mathrm {L}^s([0,T],{\mathbb {R}}^m)$, where $w^{u,M}_v \in \mathrm {AC}([0,T],{\mathbb {R}}^n)$ is the unique solution to

$$\begin{aligned} \left\{ \begin{array}{l} {\dot{w}}(t) = \nabla _x f^M(x^M_u(t),u(t),t) w(t) + \nabla _u f^M(x^M_u(t),u(t),t) v(t), \quad \text {a.e. } \ t \in [0,T] , \\ w(0) = 0_{{\mathbb {R}}^n} . \end{array} \right. \end{aligned}$$

Remark B.1

Let $1 < s \le +\infty$. For a given control $u \in {\mathcal {U}}$, note that $\mathrm {D}\mathrm {E}(u)$ given in (1) admits a natural extension (still denoted by) $\mathrm {D}\mathrm {E}(u):\mathrm {L}^s([0,T],{\mathbb {R}}^m)\rightarrow {\mathbb {R}}^n$. The nontruncated setting is related to the truncated one as follows:

• (i) Let $u \in {\mathcal {U}}$ and $M > 0$ be such that $\Vert x_u \Vert _{\mathrm {C}} \le M$ and $\Vert u \Vert _{\mathrm {L}^\infty } \le M$. Then $x^M_u = x_u$ and $\mathrm {D}\mathrm {E}^M (u) = \mathrm {D}\mathrm {E}(u)$ when considering the above extension of $\mathrm {D}\mathrm {E}(u)$.

• (ii) Let $u \in \mathrm {L}^\infty ([0,T],{\mathbb {R}}^m)$. If there exists $M > 0$ such that $\Vert x^M_u \Vert _{\mathrm {C}} \le M$ and $\Vert u \Vert _{\mathrm {L}^\infty } \le M$, then $u \in {\mathcal {U}}$ and $x_u = x^M_u$.

Remark B.2

Let $M > 0$. In the case $s=1$, it can be proved that the truncated end-point mapping $\mathrm {E}^M:\mathrm {L}^1([0,T],{\mathbb {R}}^m)\rightarrow {\mathbb {R}}^n$ is Gateaux-differentiable and its Gateaux differential is given by (8). However, it is not Fréchet-differentiable (and thus not of class $\mathrm {C}^1$) in general, as shown in the next example.

Example B.2

Take $T=n=m=1$, $\mathrm {U}= {\mathbb {R}}$ and $f(x,u,t) = u^2$ for all $(x,u,t) \in {\mathbb {R}}\times {\mathbb {R}}\times [0,T]$. Consider the starting point $x^0 = 0$ and the constant control $u \equiv 0$. In that context, with $M=s=1$, it is clear that $x^M_u \equiv 0$ and that the Gateaux differential $\mathrm {D}^{\mathrm {G}} \mathrm {E}^M(u) :\mathrm {L}^1([0,T],{\mathbb {R}}^m)\rightarrow {\mathbb {R}}^n$ of the truncated end-point mapping $\mathrm {E}^M:\mathrm {L}^1([0,T],{\mathbb {R}}^m)\rightarrow {\mathbb {R}}^n$ at u, given by the expression (8), is null. Now, taking the needle-like control variation $u^\alpha _{(0,1)}$, as defined in (2), associated with the pair $(0,1) \in {\mathcal {L}}( f_u) \times \mathrm {U}$, we obtain that

$$\begin{aligned} \lim \limits _{\alpha \rightarrow 0^+} \dfrac{ \mathrm {E}^M( u + u^\alpha _{(0,1)} ) - \mathrm {E}^M (u) - \mathrm {D}^{\mathrm {G}} \mathrm {E}^M(u) \cdot u^\alpha _{(0,1)} }{ \Vert u^\alpha _{(0,1)} \Vert _{\mathrm {L}^1} } = 1. \end{aligned}$$

Therefore, $\mathrm {E}^M:\mathrm {L}^1([0,T],{\mathbb {R}}^m)\rightarrow {\mathbb {R}}^n$ is not Fréchet-differentiable at u. A similar example can be found in [29, Remark 3.1].