[proofplan] The argument has two parts. First, the autonomous Pontryagin equations imply that the maximized Hamiltonian is constant on each smooth control arc, because the explicit time derivative is absent and the state, adjoint, and stationarity equations cancel all remaining terms. The assumed matching of one-sided Hamiltonian limits then makes this constant the same across switching times. Finally, the free-final-time transversality condition for a fixed terminal manifold and no explicit terminal-time cost gives the left terminal Hamiltonian limit $\lim_{t \uparrow t_1} h(t)=0$, and constancy propagates this value along the whole extremal. [/proofplan]

[step:Show that the maximized Hamiltonian is constant on each smooth control arc]Let $I \subset [t_0,t_1]$ be a smooth control arc on which $u^*$ is regular. Define \begin{align*} h_I: I &\to \mathbb{R} \end{align*} \begin{align*} t &\mapsto H(x^*(t),p(t),u^*(t)). \end{align*} By hypothesis, $h_I$ has an absolutely continuous representative on $I$. On this arc, the Pontryagin equations are \begin{align*} \dot{x}^*(t) = \partial_p H(x^*(t),p(t),u^*(t)) \end{align*} and \begin{align*} \dot{p}(t) = -\partial_x H(x^*(t),p(t),u^*(t)) \end{align*} for $\mathcal{L}^1$-a.e. $t \in I$. By the interior stationarity hypothesis on the smooth control arc, \begin{align*} \partial_u H(x^*(t),p(t),u^*(t)) = 0 \end{align*} for $\mathcal{L}^1$-a.e. $t \in I$. For such $t$, the chain rule for absolutely continuous compositions gives \begin{align*} \frac{d}{dt} h_I(t) = \partial_x H(x^*(t),p(t),u^*(t)) \cdot \dot{x}^*(t) + \partial_p H(x^*(t),p(t),u^*(t)) \cdot \dot{p}(t) + \partial_u H(x^*(t),p(t),u^*(t)) \cdot \dot{u}^*(t). \end{align*} Substituting the state equation, the adjoint equation, and the stationarity condition yields \begin{align*} \frac{d}{dt} h_I(t) = \partial_x H(x^*(t),p(t),u^*(t)) \cdot \partial_p H(x^*(t),p(t),u^*(t)) - \partial_p H(x^*(t),p(t),u^*(t)) \cdot \partial_x H(x^*(t),p(t),u^*(t)). \end{align*} The two scalar products are equal, so \begin{align*} \frac{d}{dt} h_I(t) = 0 \end{align*} for $\mathcal{L}^1$-a.e. $t \in I$. Since $h_I$ is absolutely continuous and has derivative zero almost everywhere, $h_I$ is constant on $I$.[/step]

[guided]Fix one smooth control arc $I \subset [t_0,t_1]$. The purpose of restricting to such an arc is that the usual differential form of the Pontryagin equations and the stationarity condition are valid there. Define \begin{align*} h_I: I &\to \mathbb{R} \end{align*} \begin{align*} t &\mapsto H(x^*(t),p(t),u^*(t)). \end{align*} The hypothesis that $h_I$ has an absolutely continuous representative is what allows us to prove constancy by proving that its derivative vanishes almost everywhere. On $I$, the state equation says \begin{align*} \dot{x}^*(t) = \partial_p H(x^*(t),p(t),u^*(t)) \end{align*} for $\mathcal{L}^1$-a.e. $t \in I$. The adjoint equation says \begin{align*} \dot{p}(t) = -\partial_x H(x^*(t),p(t),u^*(t)) \end{align*} for $\mathcal{L}^1$-a.e. $t \in I$. The repaired theorem statement includes the interior stationarity condition on each smooth control arc, so \begin{align*} \partial_u H(x^*(t),p(t),u^*(t)) = 0 \end{align*} for $\mathcal{L}^1$-a.e. $t \in I$. Now apply the chain rule to the absolutely continuous map $h_I$. For $\mathcal{L}^1$-a.e. $t \in I$, \begin{align*} \frac{d}{dt} h_I(t) = \partial_x H(x^*(t),p(t),u^*(t)) \cdot \dot{x}^*(t) + \partial_p H(x^*(t),p(t),u^*(t)) \cdot \dot{p}(t) + \partial_u H(x^*(t),p(t),u^*(t)) \cdot \dot{u}^*(t). \end{align*} The stationarity condition eliminates the final term. Substituting the state and adjoint equations into the first two terms gives \begin{align*} \frac{d}{dt} h_I(t) = \partial_x H(x^*(t),p(t),u^*(t)) \cdot \partial_p H(x^*(t),p(t),u^*(t)) - \partial_p H(x^*(t),p(t),u^*(t)) \cdot \partial_x H(x^*(t),p(t),u^*(t)). \end{align*} These are the same real scalar product with opposite signs, hence \begin{align*} \frac{d}{dt} h_I(t) = 0 \end{align*} for $\mathcal{L}^1$-a.e. $t \in I$. An absolutely continuous real-valued function whose derivative is zero almost everywhere is constant, so $h_I$ is constant on the smooth control arc $I$.[/guided]

[step:Identify the constants across switching times] Let $\tau \in (t_0,t_1)$ be a switching time between two adjacent smooth control arcs $I_-$ and $I_+$. By the previous step, there exist constants $c_-,c_+ \in \mathbb{R}$ such that \begin{align*} h(t)=c_- \end{align*} for $\mathcal{L}^1$-a.e. $t \in I_-$, and \begin{align*} h(t)=c_+ \end{align*} for $\mathcal{L}^1$-a.e. $t \in I_+$. The assumed matching of one-sided limits gives \begin{align*} \lim_{t \uparrow \tau} h(t) = \lim_{t \downarrow \tau} h(t). \end{align*} Since the left and right limits are respectively $c_-$ and $c_+$, we get $c_- = c_+$. Applying this argument at every switching time shows that there is a single constant $c \in \mathbb{R}$ such that \begin{align*} H(x^*(t),p(t),u^*(t)) = c \end{align*} for $\mathcal{L}^1$-a.e. $t \in [t_0,t_1]$. [/step]

[step:Use free-final-time transversality to compute the common constant] By the free-final-time transversality hypothesis in the theorem statement, the endpoint condition for a fixed terminal manifold $M$ and terminal cost with no explicit dependence on $t_1$ is \begin{align*} \lim_{t \uparrow t_1} h(t) = 0. \end{align*} Here the terminal manifold is time-independent by hypothesis, so there is no additional terminal-manifold time-velocity term, and the absence of explicit terminal-time cost removes the terminal-cost derivative term. Since the previous step shows that the one-sided representative of $h$ has the common constant value $c$ on the final smooth control arc, this left terminal limit equals $c$. Hence \begin{align*} c=0. \end{align*} [/step]

[step:Propagate the terminal value along the whole extremal] From the preceding steps, there exists a constant $c \in \mathbb{R}$ such that \begin{align*} H(x^*(t),p(t),u^*(t)) = c \end{align*} for $\mathcal{L}^1$-a.e. $t \in [t_0,t_1]$, and the free-final-time transversality condition in left-limit form gives $c=0$. Therefore \begin{align*} H(x^*(t),p(t),u^*(t)) = 0 \end{align*} for $\mathcal{L}^1$-a.e. $t \in [t_0,t_1]$, which is the desired conclusion. [/step]