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Consider the deterministic controlled system:

$$\dot x(t) = Ax(t) + Bu(t), \ t \in [0, T]$$ $$x(0) = x_0$$

where $x: [0, T] \to \mathbb R^n$ is the controlled state process, $A \in \mathbb R^{n \times n}$, $B \in \mathbb R^{n \times k}$ are real valued matrices, $u: [0, T] \to \mathbb R^k$ is the control process, and $x_0 \in \mathbb R^n$ is an arbitrary fixed initial point.

For each $U \subset \mathbb R^k$, we define the reachable set $\mathcal R_U$ as the set of all possible final states of the system using controls with values in $U$ - that is, the set

$$\mathcal R_U := \{x(T, u(.)) \ | \ u(t) \in U, \forall t \in [0, T]\}$$

In the book Stochastic Controls by Yong and Zhao, the following proposition is stated as Proposition 6.1 on page 77:

Proposition: Let $U \subset \mathbb R^k$ be compact. Then $\mathcal R_U$ is convex and compact in $\mathbb R^n$, and further it is equal to $\mathcal R_{\overline{\text{co } U}}$, where $\overline{\text{co }U}$ denotes the closure of the convex hull of $U$.

The proposition is said in turn to be a consequence of the following Lyapunov’s theorem, which is Theorem 6.3 on the same page:

Theorem: Suppose $f \in L^1 ([0, T], \mathbb R^n)$. Then the set $\mathcal H := \{\int_S f(x) dx \ | \ S \in \mathcal B [0, T]\}$ is a convex subset of $\mathbb R^n$.

where here $\mathcal B[0, T]$ is the set of Borel subsets of $[0, T]$.

On the same page, they refer to the book Functional Analysis and Time Optimal Controls by Hermes and LaSalle for proofs of both these statements. However, I was only able to find a proof of the Theorem above and not the Proposition.

Question:

Does anyone know of an alternative reference for this result?

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  • $\begingroup$ Title of your post is misleading.Your post says that the $\text{reachable sets}$ form a convex set when control is in a compact set. $\endgroup$ Commented Oct 3, 2021 at 3:06
  • $\begingroup$ @PiyushGrover It refers to the second part of the proposition that states that $\mathcal R_U$ is equal to $\mathcal R_{\overline {\text{co U}}}$. $\endgroup$
    – Nate River
    Commented Oct 3, 2021 at 3:08
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    $\begingroup$ The reason why the Proposition follows from the Theorem is this: the set $\mathcal{R}_{U}$ can be written as the translation of a set-valued integral given by $\exp(tA)x_{0} + \int_{0}^{t}\exp(sA)B U\:ds$. The plus sign denotes a Minkowski sum. The set valued integral, by Lyapunov's theorem, is a convex set and compact. From here, you can also show that the set-valued integral is invariant under the closure of convexification of $U$. $\endgroup$ Commented Oct 3, 2021 at 6:12
  • $\begingroup$ Thank you! Can you possibly elaborate a litle more on how Lyapunov’s theorem implies the set valued integral is convex? As stated it applies to a single function integrated over all possible Borel sets, while your version is over different points in $U$. It is not immediately obvious to me how to go from my version to yours. $\endgroup$
    – Nate River
    Commented Oct 3, 2021 at 6:39
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    $\begingroup$ I will leave the formalization but mention the key ideas. In the set-valued integral I wrote, the integrand is a set parameterized by a choice of $s$ in $[0,t]$. The set $U$ is fixed. Different choices of $s$ in the integrand result in different linear transforms of the set $U$. If you uniformly discretize $[0,t]$, then that integral is the limit of the Minkowski sum of these linearly transformed sets, where the limit is w.r.t. discretization length. So Lyapunov theorem is saying that repeated Minkowski sums eventually convexify. You can view it as the continuum version of Shapley-Folkman. $\endgroup$ Commented Oct 6, 2021 at 21:05

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