[proofplan] We take a minimizing sequence and use compactness of the state and control sets to obtain a uniformly Lipschitz family of trajectories. The Arzela-Ascoli theorem then gives [uniform convergence](/page/Uniform%20Convergence) of a subsequence to an absolutely continuous trajectory with values in $K$. The decisive step is the Filippov-Cesari closure theorem: convexity of the velocity-cost epigraphs allows the weak limit of velocity-cost pairs to be represented by an actual measurable control for the limiting trajectory. Closedness of the endpoint constraint and lower semicontinuity of the terminal cost then pass feasibility and optimality to the limit. [/proofplan] [step:Reduce to a finite minimizing sequence] Let $\mathcal{A}$ denote the set of admissible pairs. By hypothesis, $\mathcal{A}$ is nonempty. Define \begin{align*} \alpha := \inf\{J[x,u] : (x,u) \in \mathcal{A}\}. \end{align*} Since $J$ is bounded below on $\mathcal{A}$, $\alpha > -\infty$. If $\alpha = \infty$, then every admissible pair has value $\infty$, so any element of $\mathcal{A}$ is optimal. Hence assume from now on that $\alpha < \infty$. Choose a sequence $(x_k,u_k)_{k \in \mathbb{N}}$ in $\mathcal{A}$ such that \begin{align*} J[x_k,u_k] \to \alpha. \end{align*} Discarding finitely many terms if necessary, we assume $J[x_k,u_k] < \infty$ for every $k \in \mathbb{N}$; this is possible because $\alpha \in \mathbb{R}$ in the present case. In particular, $\Phi(x_k(t_1)) < \infty$ for every $k \in \mathbb{N}$. For each $k \in \mathbb{N}$, define the running-cost function $\ell_k: [t_0,t_1] \to \mathbb{R}$ by \begin{align*} \ell_k(t) := L(t,x_k(t),u_k(t)). \end{align*} The map $\ell_k$ is Lebesgue measurable because $u_k$ is Lebesgue measurable, $x_k$ is continuous, and $L$ is continuous. [/step] [step:Extract a uniformly convergent subsequence of trajectories] Since $K$ is compact, define the finite radius constant \begin{align*} R_K := \max\{|z| : z \in K\}. \end{align*} Using the stated linear growth bound for $f$, define \begin{align*} M_f := a + bR_K. \end{align*} For every $k \in \mathbb{N}$ and for $\mathcal{L}^1$-a.e. $t \in [t_0,t_1]$, the inclusion $x_k(t) \in K$ gives $|x_k(t)| \leq R_K$, and therefore \begin{align*} |\dot{x}_k(t)| = |f(t,x_k(t),u_k(t))| \leq a + b|x_k(t)| \leq M_f. \end{align*} Thus each $x_k$ is $M_f$-Lipschitz: for all $s,t \in [t_0,t_1]$, \begin{align*} |x_k(t)-x_k(s)| \leq M_f |t-s|. \end{align*} The sequence $(x_k)_{k \in \mathbb{N}}$ is therefore uniformly bounded, because $x_k([t_0,t_1]) \subset K$, and equicontinuous. By the Arzela-Ascoli theorem, after passing to a subsequence without changing notation, there exists a continuous map $x: [t_0,t_1] \to \mathbb{R}^n$ such that $x_k \to x$ uniformly on $[t_0,t_1]$. Since $K$ is compact and hence closed, and since $x_k(t) \in K$ for every $k$ and every $t$, we have $x(t) \in K$ for every $t \in [t_0,t_1]$. Passing to the limit in the Lipschitz estimate gives \begin{align*} |x(t)-x(s)| \leq M_f |t-s| \end{align*} for all $s,t \in [t_0,t_1]$, so $x$ is Lipschitz and therefore absolutely continuous. [/step] [step:Apply Filippov-Cesari closure to recover a limiting control] For each $k \in \mathbb{N}$, define the velocity-cost map $y_k: [t_0,t_1] \to \mathbb{R}^{n+1}$ by \begin{align*} y_k(t) := (\dot{x}_k(t),\ell_k(t)) \end{align*} where $\dot{x}_k(t)$ is defined. Since $(x_k,u_k)$ is admissible, for $\mathcal{L}^1$-a.e. $t \in [t_0,t_1]$, \begin{align*} y_k(t) = (f(t,x_k(t),u_k(t)),L(t,x_k(t),u_k(t))) \in Q(t,x_k(t)). \end{align*} Also, \begin{align*} x_k(t) = x_k(t_0) + \int_{[t_0,t]} \dot{x}_k(s)\,d\mathcal{L}^1(s). \end{align*} We now apply the Filippov-Cesari closure theorem for convex velocity-cost epigraphs. Its hypotheses are satisfied. First, $x_k \to x$ uniformly and each $x_k$ takes values in the compact set $K$. Second, the multifunction $(t,z) \mapsto Q(t,z)$ has nonempty convex values by the nonemptiness of $U_c$ and the stated convexity assumption. Its values are closed: if $(v_j,r_j) \in Q(t,z)$ and $(v_j,r_j) \to (v,r)$ in $\mathbb{R}^{n+1}$, choose $u_j \in U_c$ with $v_j=f(t,z,u_j)$ and $r_j \geq L(t,z,u_j)$; compactness of $U_c$ gives a subsequence $u_{j_i} \to u \in U_c$, and continuity of $f$ and $L$ gives $v=f(t,z,u)$ and $r \geq L(t,z,u)$. Third, the graph of $Q$ is closed in $[t_0,t_1] \times K \times \mathbb{R}^{n+1}$. Indeed, suppose $(t_j,z_j,v_j,r_j) \to (t,z,v,r)$ in $[t_0,t_1] \times K \times \mathbb{R}^{n+1}$ and $(v_j,r_j) \in Q(t_j,z_j)$ for every $j \in \mathbb{N}$. Choose $u_j \in U_c$ such that $v_j=f(t_j,z_j,u_j)$ and $r_j \geq L(t_j,z_j,u_j)$. Compactness of $U_c$ gives a subsequence $u_{j_i} \to u \in U_c$. Continuity of $f$ and $L$ gives $v=f(t,z,u)$ and $r \geq L(t,z,u)$, hence $(v,r) \in Q(t,z)$. Therefore the required measurability and upper-closedness hypotheses hold for this continuous compact-control epigraph multifunction. Fourth, the velocity components are uniformly bounded by $M_f$ by the preceding step. The cost coordinates are uniformly bounded below and above because continuity of $L$ on the compact set $[t_0,t_1] \times K \times U_c$ gives constants $m_L,M_L \in \mathbb{R}$ such that $m_L \leq L(t,z,w) \leq M_L$ for all $(t,z,w)$ in that compact set; hence the constant functions $t \mapsto m_L$ and $t \mapsto M_L$ belong to $L^1([t_0,t_1])$. Therefore there exist a Lebesgue measurable control $u: [t_0,t_1] \to U_c$ and a function $\ell \in L^1([t_0,t_1])$ such that, for $\mathcal{L}^1$-a.e. $t \in [t_0,t_1]$, \begin{align*} \dot{x}(t) = f(t,x(t),u(t)) \end{align*} and \begin{align*} \ell(t) \geq L(t,x(t),u(t)). \end{align*} Moreover the closure theorem gives the cost lower-closure inequality \begin{align*} \int_{[t_0,t_1]} \ell(t)\,d\mathcal{L}^1(t) \leq \liminf_{k \to \infty} \int_{[t_0,t_1]} \ell_k(t)\,d\mathcal{L}^1(t). \end{align*} [guided] The purpose of the convex epigraph hypothesis is to make the limit of the velocity-cost pairs representable by an actual control. For each $k$, define \begin{align*} y_k(t) := (\dot{x}_k(t),\ell_k(t)). \end{align*} Because $(x_k,u_k)$ satisfies the dynamics and because $\ell_k(t)=L(t,x_k(t),u_k(t))$, we have, for $\mathcal{L}^1$-a.e. $t$, \begin{align*} y_k(t) = (f(t,x_k(t),u_k(t)),L(t,x_k(t),u_k(t))) \in Q(t,x_k(t)). \end{align*} Thus $y_k(t)$ lies on the graph of the velocity-cost epigraph associated to the point $(t,x_k(t))$. The limiting difficulty is that the controls $u_k$ need not converge pointwise or weakly to a meaningful control. The Filippov-Cesari closure theorem is designed precisely for this situation. It says that, when the velocity-cost epigraphs are closed and convex, a uniform limit of trajectories whose velocity-cost pairs lie in those epigraphs is again generated by a measurable control, with the running cost controlled by a liminf inequality. We verify the hypotheses. The convergence $x_k \to x$ is uniform by the preceding step. The values $x_k(t)$ and $x(t)$ lie in the compact set $K$. For each fixed $(t,z) \in [t_0,t_1] \times K$, the set $Q(t,z)$ is nonempty because $U_c$ is nonempty, and it is convex by hypothesis. We next check closedness of the epigraph values. Suppose $(v_j,r_j) \in Q(t,z)$ and $(v_j,r_j) \to (v,r)$ in $\mathbb{R}^{n+1}$. For each $j$, choose $w_j \in U_c$ such that $v_j=f(t,z,w_j)$ and $r_j \geq L(t,z,w_j)$. Since $U_c$ is compact, a subsequence satisfies $w_{j_i} \to w$ for some $w \in U_c$. Continuity of $f$ and $L$ gives $v=f(t,z,w)$ and $r \geq L(t,z,w)$, so $(v,r) \in Q(t,z)$. Thus each $Q(t,z)$ is closed. We verify the closed-graph condition required by the closure theorem directly. Suppose $(t_j,z_j,v_j,r_j) \to (t,z,v,r)$ in $[t_0,t_1] \times K \times \mathbb{R}^{n+1}$ and $(v_j,r_j) \in Q(t_j,z_j)$ for every $j \in \mathbb{N}$. By the definition of $Q(t_j,z_j)$, choose $w_j \in U_c$ with $v_j=f(t_j,z_j,w_j)$ and $r_j \geq L(t_j,z_j,w_j)$. Since $U_c$ is compact, there is a subsequence $w_{j_i} \to w$ for some $w \in U_c$. Continuity of $f$ and $L$ gives $v=f(t,z,w)$ and $r \geq L(t,z,w)$. Hence $(v,r) \in Q(t,z)$, so the graph is closed. For this compact-control continuous epigraph multifunction, closed graph gives the measurability and upper-closedness assumptions used in the Filippov-Cesari closure theorem. Finally, the velocity components are uniformly bounded because the preceding step proved \begin{align*} |\dot{x}_k(t)| = |f(t,x_k(t),u_k(t))| \leq M_f \end{align*} for $\mathcal{L}^1$-a.e. $t$. The cost coordinates have integrable lower and upper bounds: continuity of $L$ on the compact set $[t_0,t_1] \times K \times U_c$ gives constants $m_L,M_L \in \mathbb{R}$ such that $m_L \leq L(t,z,w) \leq M_L$ for all $(t,z,w)$ in that compact set, and the constant functions $t \mapsto m_L$ and $t \mapsto M_L$ belong to $L^1([t_0,t_1])$. Applying the Filippov-Cesari closure theorem gives a Lebesgue measurable map $u: [t_0,t_1] \to U_c$ and an integrable function $\ell: [t_0,t_1] \to \mathbb{R}$ such that, for $\mathcal{L}^1$-a.e. $t$, \begin{align*} \dot{x}(t) = f(t,x(t),u(t)). \end{align*} The same closure result places the limiting cost coordinate above the selected running cost: \begin{align*} \ell(t) \geq L(t,x(t),u(t)) \end{align*} for $\mathcal{L}^1$-a.e. $t$. It also gives the essential lower-closure estimate \begin{align*} \int_{[t_0,t_1]} \ell(t)\,d\mathcal{L}^1(t) \leq \liminf_{k \to \infty} \int_{[t_0,t_1]} \ell_k(t)\,d\mathcal{L}^1(t). \end{align*} This is the point where convexity is used: without convex velocity-cost epigraphs, weak limits of rapidly oscillating controls may produce averaged velocity-cost pairs that do not come from any single admissible control value. [/guided] [/step] [step:Pass endpoint feasibility to the limit] Since $x_k \to x$ uniformly, in particular \begin{align*} x_k(t_0) \to x(t_0) \end{align*} and \begin{align*} x_k(t_1) \to x(t_1). \end{align*} For every $k \in \mathbb{N}$, admissibility gives $(x_k(t_0),x_k(t_1)) \in E$. Since $E$ is closed in $\mathbb{R}^n \times \mathbb{R}^n$, it follows that \begin{align*} (x(t_0),x(t_1)) \in E. \end{align*} Together with $x([t_0,t_1]) \subset K$, absolute continuity of $x$, measurability of $u$, and the a.e. dynamics obtained above, this proves that $(x,u)$ is admissible. [/step] [step:Use lower semicontinuity to prove optimality] Because $\Phi$ is lower semicontinuous on $K$ and $x_k(t_1) \to x(t_1)$ in $K$, \begin{align*} \Phi(x(t_1)) \leq \liminf_{k \to \infty} \Phi(x_k(t_1)). \end{align*} Since $\ell(t) \geq L(t,x(t),u(t))$ for $\mathcal{L}^1$-a.e. $t$, monotonicity of the [Lebesgue integral](/page/Lebesgue%20Integral) gives \begin{align*} \int_{[t_0,t_1]} L(t,x(t),u(t))\,d\mathcal{L}^1(t) \leq \int_{[t_0,t_1]} \ell(t)\,d\mathcal{L}^1(t). \end{align*} Combining this with the Filippov-Cesari liminf inequality yields \begin{align*} \int_{[t_0,t_1]} L(t,x(t),u(t))\,d\mathcal{L}^1(t) \leq \liminf_{k \to \infty} \int_{[t_0,t_1]} L(t,x_k(t),u_k(t))\,d\mathcal{L}^1(t). \end{align*} The function $L$ is continuous on the compact set $[t_0,t_1] \times K \times U_c$, so the running-cost integrals are bounded below by a finite constant. Since we have discarded to a finite-value minimizing sequence, each $\Phi(x_k(t_1))$ is a real number. The terminal values in the minimizing sequence are bounded below because $J[x_k,u_k] \to \alpha \in \mathbb{R}$ and the running costs are bounded. Hence the standard superadditivity of $\liminf$ for bounded-below real sequences gives \begin{align*} J[x,u] \leq \liminf_{k \to \infty} J[x_k,u_k]. \end{align*} Since $J[x_k,u_k] \to \alpha$, we obtain \begin{align*} J[x,u] \leq \alpha. \end{align*} On the other hand, $(x,u)$ is admissible, so the definition of $\alpha$ gives \begin{align*} \alpha \leq J[x,u]. \end{align*} Therefore \begin{align*} J[x,u] = \alpha. \end{align*} Thus $(x,u)$ is an optimal admissible pair. [/step]