The Set of Controls with Values in a Compact Convex Set is Weakly Metrizable and Compact

Let $T>0$ be a real number and $m$ a natural number, write $H=L^{2}([0,T];\mathbb{R}^{m})$ and adopt the pairing $\langle\cdot,\cdot\rangle_{L^{2}}$, the norm $\lVert\cdot\rVert_{L^{2}}$ and the metric $d_{L^{2}}$ of the Lebesgue space in the case $d=m$.

Let $\mathcal{A}$ be a nonempty subset of $\mathbb{R}^{m}$ that is compact for the topology determined by the Euclidean distance and satisfies $sa+(1-s)a'\in\mathcal{A}$ for all $a,a'\in\mathcal{A}$ and all real $s$ with $0\le s\le1$, and let $\mathcal{U}_{\mathcal{A}}$ be the set of $\mathcal{A}$-valued controls.

Let $(w_{r})_{r\in\mathbb{N}}$ be a sequence in $H$ whose set of terms is dense in the metric space $(H,d_{L^{2}})$; such a sequence exists because Separability of the Lebesgue Space of Square-Integrable Vector-Valued Functions provides a dense subset of $(H,d_{L^{2}})$ that is countable, and a countable set that is nonempty is the set of terms of a sequence. For $\xi,\zeta\in H$ put

where $2^{-r}$ is formed with the natural powers of $\mathbb{R}$ and $|\cdot|$ is the absolute value. Then the following hold.

1. (A metric.) The supremum defining $\rho(\xi,\zeta)$ exists for all $\xi,\zeta\in H$, and $\rho$ is a metric on $H$. In particular the restriction of $\rho$ to $\mathcal{U}_{\mathcal{A}}$ makes $\mathcal{U}_{\mathcal{A}}$ a metric space.

2. ($\rho$ metrizes weak convergence on $\mathcal{U}_{\mathcal{A}}$.) Let $(\xi_{n})_{n\in\mathbb{N}}$ be a sequence in $\mathcal{U}_{\mathcal{A}}$ and let $\xi\in\mathcal{U}_{\mathcal{A}}$. Then $\xi_{n}\rightharpoonup\xi$ in the sense of weak convergence if and only if the real sequence $\bigl(\rho(\xi_{n},\xi)\bigr)_{n\in\mathbb{N}}$ has limit $0$.

3. (Compactness.) $\mathcal{U}_{\mathcal{A}}$ is a sequentially compact subset of the metric space $(\mathcal{U}_{\mathcal{A}},\rho)$, and hence a compact subset of it.