Let \(\mathcal{X}\) be an open subset of \(\mathbb{K}^{n}\) and let \(\mathcal{U}\) be a metric space. \(A\) right-hand side (rhs) with respect to \(X\) and \(\mathcal{U}\) is a function $$ f: \mathbb{R} \times \mathcal{X} \times \mathcal{U} \rightarrow \mathbb{K}^{n} $$ which can be obtained in the following way: There must exist another metric space \(\mathcal{S}\) as well as maps $$ \tilde{f}: \mathcal{S} \times x \times \mathcal{U} \rightarrow \mathbb{K}^{n} $$ and $$ \pi: \mathbb{R} \rightarrow \mathcal{S} $$ so that $$ f(t, x, u)=\tilde{f}(\pi(t), x, u) $$ and the following properties hold: 1\. \(\widetilde{f}(s, \cdot, u)\) is of class \(\mathcal{C}^{1}\) for each fixed \(s, u\); 2\. both \(\tilde{f}\) and the partial derivative \(\tilde{f}_{x}\) are continuous on \(\mathcal{S} \times \mathcal{X} \times \mathcal{U}\); 3\. \(\pi\) is a measurable locally essentially bounded function.

Let \(m, n\) be nonnegative integers, $x:=\mathbb{K}^{n}, \mathcal{U}:=\mathbb{K}^{m} .$ Assume that $$ f(t, x, u)=A(t) x+B(t) u $$ where \(A(t)\) is an \(n \times n\) matrix and \(B(t)\) is an \(n \times m\) matrix, each of whose entries is a locally essentially bounded (measurable) function \(\mathbb{R} \rightarrow \mathbb{K}\). Equivalently, $$ f: \mathbb{R} \times \mathcal{X} \times \mathcal{U} \rightarrow \mathcal{X} $$ is linear in \((x, u)\) for each fixed \(t\), and is locally bounded in \(t\) for each fixed \((x, u)\). Conclude that \(f\) is a rhs.

Any rhs in the sense of Definition 2.6.1 satisfies the following property: For any real numbers \(\sigma<\tau\), any measurable essentially bounded $\omega \in \mathcal{U}^{[\sigma, \tau)}\(, and any \)x^{0} \in \mathcal{X}$, there is some nonempty subinterval \(J \subseteq \mathcal{I}:=[\sigma, \tau]\) open relative to \(\mathcal{I}\) and containing \(\sigma\), and there exists a solution of $$ \begin{aligned} \dot{\xi}(t) &=f(t, \xi(t), \omega(t)) \\ \xi(\sigma) &=x^{0} \end{aligned} $$ on \(J\). Furthermore, this solution is maximal and unique: that is, if $$ \zeta: J^{\prime} \rightarrow X $$ is any other solution of \((2.24)\) defined on a subinterval $J^{\prime} \subseteq \mathcal{I}\(, then necessarily \)J^{\prime} \subseteq J$ and \(\xi=\zeta\) on \(J^{\prime} .\) If \(J=\mathcal{I}\), then \(\omega\) is said to be admissible for \(x^{0} .\)

If \(\Sigma\) is time-invariant, then \(\Lambda_{\Sigma}\) also is.

Assume that \(K\) in formula (2.27) is continuous. Pick any fixed $i \in\\{1, \ldots, m\\}\( and any \)T>0\(. For each \)\varepsilon>0\(, let \)\omega_{e}$ be the control defined on the interval \([0, T]\) by $$ \omega_{\varepsilon}(t):= \begin{cases}\frac{1}{\varepsilon} e_{i} & \text { if } 0 \leq t<\varepsilon \\ 0 & \text { otherwise }\end{cases} $$ where \(e_{i}=(0, \ldots, 0,1,0, \ldots, 0)^{\prime}\) is the \(i\) th canonical basis vector. Prove that $\lim _{\varepsilon \rightarrow 0} y_{\varepsilon}(T)=K_{i}(T)$,

A continuous-time behavior is one for which \- \(\mathcal{T}=\mathbb{R}\); \- U is a metric space; \- \(y\) is a metric space; and for each \(\sigma<\tau\) it holds that the domain of each $\lambda^{\sigma, \tau}\( is an open subset of \)\mathcal{L}_{U}^{\infty}(\sigma, \tau)$ and \(\lambda^{\sigma, \tau}\) is continuous.