[proofplan] The result is the free-endpoint specialization of the fixed-time Pontryagin maximum principle with endpoint cost. The maximum principle supplies an absolutely continuous adjoint arc and the adjoint differential equation under the stated normal normalization $p_0=-1$. The only point to identify is the terminal condition: because the terminal endpoint is unconstrained, its admissible first-order variations fill all of $\mathbb{R}^n$, so the terminal covector $p(t_1)+\nabla\Phi(x^*(t_1))$ must annihilate every vector in $\mathbb{R}^n$. Hence that covector is zero. [/proofplan] [step:Record the normal adjoint equation supplied by the extremal hypothesis] By the definition of admissible normal extremal used in the statement, with cost multiplier $p_0=-1$ and Hamiltonian \begin{align*} H(t,x,u,p)=p\cdot f(t,x,u)-L(t,x,u), \end{align*} there is an adjoint arc $p:[t_0,t_1]\to\mathbb{R}^n$ that is absolutely continuous and satisfies \begin{align*} \dot{p}(t)=-\frac{\partial H}{\partial x}(t,x^*(t),u^*(t),p(t)) \end{align*} for $\mathcal{L}^1$-a.e. $t\in[t_0,t_1]$. The same hypothesis also supplies the endpoint transversality relation that $p(t_1)+\nabla\Phi(x^*(t_1))$ annihilates every admissible first-order terminal variation. [guided] The theorem is not proving the full Pontryagin maximum principle. Instead, the phrase "admissible normal extremal" has been made explicit in the statement: it means that, under the normal normalization $p_0=-1$, the associated adjoint arc satisfies the fixed-time normal adjoint equation and the endpoint transversality relation. With this convention the Hamiltonian is \begin{align*} H(t,x,u,p)=p\cdot f(t,x,u)-L(t,x,u). \end{align*} The regularity assumptions that $f$ and $L$ are continuously differentiable in the state variable ensure that $\frac{\partial H}{\partial x}(t,x^*(t),u^*(t),p(t))$ is defined along the extremal, and $\Phi\in C^1(\mathbb{R}^n)$ ensures that $\nabla\Phi(x^*(t_1))$ is defined. Thus the extremal hypothesis gives an absolutely continuous map $p:[t_0,t_1]\to\mathbb{R}^n$ satisfying \begin{align*} \dot{p}(t)=-\frac{\partial H}{\partial x}(t,x^*(t),u^*(t),p(t)) \end{align*} for $\mathcal{L}^1$-a.e. $t\in[t_0,t_1]$. It also gives the endpoint condition before specializing to the free-endpoint case: the covector $p(t_1)+\nabla\Phi(x^*(t_1))$ vanishes on every admissible first-order terminal variation. The remaining work is to identify that variation space when there is no terminal constraint. [/guided] [/step] [step:Use the absence of terminal constraints to identify the terminal normal space] Let $y_1:=x^*(t_1)\in\mathbb{R}^n$ denote the terminal point of the extremal. Since there is no terminal state constraint and no active state constraint at $t_1$, the terminal constraint set is all of $\mathbb{R}^n$. Hence the set of admissible first-order terminal variations is the tangent space \begin{align*} T_{y_1}\mathbb{R}^n=\mathbb{R}^n. \end{align*} We identify $(\mathbb{R}^n)^*$ with $\mathbb{R}^n$ through the Euclidean [inner product](/page/Inner%20Product), so a covector represented by $a\in\mathbb{R}^n$ acts on $\eta\in\mathbb{R}^n$ as $a\cdot\eta$. Under this identification, the annihilator of $T_{y_1}\mathbb{R}^n=\mathbb{R}^n$ is $\{0\}$. The endpoint transversality condition in the statement says that the covector represented by \begin{align*} p(t_1)+\nabla\Phi(y_1) \end{align*} annihilates every admissible terminal variation. Therefore, for every vector $\eta\in\mathbb{R}^n$, \begin{align*} \bigl(p(t_1)+\nabla\Phi(y_1)\bigr)\cdot \eta=0. \end{align*} [/step] [step:Conclude that the terminal covector vanishes] Set \begin{align*} q:=p(t_1)+\nabla\Phi(y_1)\in\mathbb{R}^n. \end{align*} The preceding step gives $q\cdot\eta=0$ for every $\eta\in\mathbb{R}^n$. Taking $\eta=q$ gives \begin{align*} |q|^2=0. \end{align*} Hence $q=0$, so \begin{align*} p(t_1)+\nabla\Phi(y_1)=0. \end{align*} Substituting $y_1=x^*(t_1)$ yields \begin{align*} p(t_1)=-\nabla\Phi(x^*(t_1)). \end{align*} Together with the absolute continuity of $p$ and the adjoint equation obtained from the maximum principle, this proves the theorem. [/step]