where $\kappa(\delta) = \sup_{1 \le i \le p} \sup_{\substack{x,y \in \mathbb{R}^p \\ |x-y| \le \delta}} |\partial_i \tilde{Q}(x) - \partial_i \tilde{Q}(y)|$. We define $\tilde{Z} = \{x \in \mathbb{R}^p : d(x, Z) \le 1\}$. Since $\partial \tilde{Q}$ is uniformly null outside $\tilde{Z}$, and assuming that $\delta \le 1$, we notice that

$$ \kappa(\delta) = \sup_{1 \le i \le p} \sup_{\substack{x,y \in \tilde{Z} \times \mathbb{R}^p \\ |x-y| \le \delta}} |\partial_i \tilde{Q}(x) - \partial_i \tilde{Q}(y)| = \sup_{1 \le i \le p} \sup_{\substack{x,y \in \tilde{Z} \times \tilde{Z} \\ |x-y| \le \delta}} |\partial_i \tilde{Q}(x) - \partial_i \tilde{Q}(y)|. $$

We note that $\partial_i \tilde{Q}(x) = \tilde{\omega}(x)\partial_i Q(x) + Q(x)\partial_i \tilde{\omega}(x)$, and therefore, $|\partial_i \tilde{Q}(x) - \partial_i \tilde{Q}(y)| = |\tilde{\omega}(x)(\partial_i Q(x) - \partial_i Q(y)) + \partial_i Q(y)(\tilde{\omega}(x) - \tilde{\omega}(y)) + Q(x)(\partial_i \tilde{\omega}(x) - \partial_i \tilde{\omega}(y)) + \partial_i \tilde{\omega}(y)(Q(x) - Q(y))|$. We use, then, the triangle inequality and the facts that (i) $\tilde{\omega}$ is $C^\infty$, therefore bounded, Lipschitz continuous, and with a Lipschitz-continuous gradient over $\tilde{Z}$ and (ii) $Q$ is continuously differentiable, therefore bounded, and Lipschitz continuous over $\tilde{Z}$; by assumption it has a Lipschitz continuous gradient over the compact set $\tilde{Z}$. We deduce that $\partial_i \tilde{Q}$ is Lipschitz continuous over $\tilde{Z}$: there exists $L_i > 0$ such that $\sup_{\substack{x,y \in \tilde{Z} \times \tilde{Z} \\ |x-y| \le \delta}} |\partial_i \tilde{Q}(x) - \partial_i \tilde{Q}(y)| \le L_i \delta$, for all $\delta \in [0, 1]$. Defining $L = \max_i L_i$, we deduce

$\kappa(\delta) \le L\delta$. Then Eq. (A.4) reads $\sup_{x \in Z} |w_d(x) - \tilde{Q}(x)| \le \frac{CL}{d}$, and Eq. (A.5) reads $\sup_{x \in Z} |\partial_i(w_d - \tilde{Q})(x)| \le \frac{CL}{d}$ for all $i \in \{1, \dots, p\}$. We also deduce that $\sup_{x \in Z} \|\nabla w_d(x) - \nabla \tilde{Q}(x)\| \le \frac{CLp}{d}$. Defining $A = CLp$, and noticing that for all $x \in Z$, $\tilde{Q}(x) = Q(x)$ and $\nabla \tilde{Q}(x) = \nabla Q(x)$, one obtains the claimed statement. $\square$

**LEMMA A.3.** Under Assumptions 1 and 2, we consider an admissible trajectory $(x(\cdot), u(\cdot))$ over $[0, t_1]$ of the minimal time control problem (2.1)-(2.4) starting from $(0, x_0)$. Then, for almost all $t \in [0, t_1]$, $f(t, x(t), u(t)) \in T_X(x(t))$.

*Proof.* We reduce to a time-invariant controlled system: we define, for any $y = (t, x) \in \mathbb{R}^{n+1}$ and $u \in \mathbb{R}^m$, $\tilde{f}(y, u) = \begin{pmatrix} 1 \\ f(y, u) \end{pmatrix}$, and the constant set-valued map $\tilde{U}(y) = U$. The control system $(\tilde{f}, \tilde{U})$ is a Marchaud control system [1, Def. 6.1.3], since: (i) the graph of $\tilde{U}$ is closed (ii) $\tilde{f}$ is continuous (iii) the velocity subsets $\{\tilde{f}(y, u) : u \in \tilde{U}(y)\}$ are convex due to Assumption 1, and (iv) the function $f$ has a linear growth since it is Lipschitz continuous, and the set-valued map are bounded, thus also has a linear growth. We define the set $C = \mathbb{R}_+ \times X$ and notice that the control $u(\cdot)$ regulates a trajectory $y(t) = \begin{pmatrix} t \\ x(t) \end{pmatrix}$ that remains in $C$, therefore according to [1, Th. 6.1.4], for all most all $t \in [0, t_1]$, $u(t) \in \{u \in \tilde{U}(y(t)) : \tilde{f}(y(t), u(t)) \in T_C(y(t))\}$. We notice that $T_C(y(t)) \subset \mathbb{R} \times T_X(x(t))$, implying that $f(y(t), u(t)) \in T_X(x(t))$. $\square$

**LEMMA A.4.** For any locally Lipschitz continuous function $F: \mathbb{R}^p \to \mathbb{R}$, and for any Lipschitz continuous curve $y: [0, T] \to \mathbb{R}^p$, the function $t \mapsto F(y(t))$ is differentiable a.e. and satisfies

$$ (A.6) \qquad \min_{g \in \partial^c F(y(t))} g^\top \dot{y}(t) \le \frac{d}{dt} (F(y(t))) \le \max_{g \in \partial^c F(y(t))} g^\top \dot{y}(t), $$

for almost all $t \in [0, T]$.

The particular functions $F$ for which these three quantities are equal are called path-differentiable in [8].

*Proof.* First, we notice that the functions $t \mapsto y(t)$ and $t \mapsto F(y(t))$ are Lipschitz continuous, therefore differentiable a.e. on $[0, T]$ due to the Rademacher theorem [20]. Hence, for almost all $t \in [0, T]$, both $y(t)$ and $F(y(t))$ are differentiable at $t$. We consider such a $t$, and we show that (A.6) holds for this particular $t$, which will prove the Lemma. Since $y$ is differentiable at $t$, $r(h) = y(t+h) - y(t) - h\dot{y}(t)$ is in $o_{h \to 0}(h)$.

Since $s \mapsto F(y(s))$ is differentiable at $t$, the following holds

$$ (A.7) \qquad \frac{d}{dt} (F(y(t))) = \lim_{h \to 0, h > 0} \frac{F(y(t+h)) - F(y(t))}{h} \\ (A.8) \qquad = \lim_{h \to 0, h > 0} \frac{F(y(t) + h\dot{y}(t) + r(h)) - F(y(t))}{h} $$

Since $r(h) = o_{h \to 0}(h)$ and $F$ is locally Lipschitz, we know that $\lim_{h \to 0, h > 0} \frac{F(y(t))-F(y(t)+r(h))}{h} = 0$. Summing this with Eq. (A.8), we deduce that
$$ (A.9) \qquad \frac{d}{dt}(F(y(t))) = \lim_{h \to 0, h > 0} \frac{F(y(t) + r(h) + h\dot{y}(t)) - F(y(t) + r(h))}{h} $$

$$ (A.10) \qquad \leq \limsup_{\substack{y' \to y(t) \\ h \to 0, h > 0}} \frac{F(y' + h\dot{y}(t)) - F(y')}{h} = F^{\circ}(y(t):\dot{y}(t)), $$

where $F^{\circ}(y; v) = \limsup_{\substack{y' \to y \\ h \to 0, h > 0}} \frac{F(y'+hv)-F(y')}{h}$ is the Clarke directional derivative of $F$ at $y \in \mathbb{R}^p$ in the direction $v \in \mathbb{R}^p$. The inequality follows from the fact that $y(t) + r(h) \to y(t)$. By property of the Clarke subdifferential [13], we also know that $F^{\circ}(y; v) = \max_{g \in \partial^c F(y)} g^\top v$. Hence, in particular,

$$ (A.11) \qquad \frac{d}{dt}(F(y(t))) \leq \max_{g \in \partial^c F(y(t))} g^\top \dot{y}(t). $$

The reasoning that proved Eq. (A.11) is also applicable to $-F$, that is also locally Lipschitz, and such that $s \mapsto (-F)(s)$ is differentiable at $t$. Therefore,

$$ (A.12) \qquad \frac{d}{dt}(-F(y(t))) \leq \max_{g \in \partial^c(-F)(y(t))} g^\top \dot{y}(t). $$

As $\partial^c(-F)(y(t)) = -\partial^c F(y(t))$ by property of the Clarke subdifferential, we deduce that

$$ (A.13) \qquad -\frac{d}{dt}(F(y(t))) \leq \max_{g \in \partial^c F(y(t))} -g^\top \dot{y}(t) = -\min_{g \in \partial^c F(y(t))} g^\top \dot{y}(t), $$

and therefore $\frac{d}{dt}(F(y(t))) \geq \min_{g \in \partial^c F(y(t))} g^\top \dot{y}(t)$. $\square$

<figure><img src="image_8.png" /></figure>