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

Preliminaries. Since $\mathcal{A}$ is compact it is closed by Compact Subset of $\mathbb{R}^n$ is Closed and bounded by Compact Subset of $\mathbb{R}^n$ is Bounded, so there are $x\in\mathbb{R}^{m}$ and a real $R'\ge0$ with $d_{E}(x,a)\le R'$ for every $a\in\mathcal{A}$, where $d_{E}$ is the Euclidean distance. Putting $R=|x|+R'$ and using the triangle inequality for the Euclidean norm (Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$) gives $|a|\le|a-x|+|x|=d_{E}(x,a)+|x|\le R$ for every $a\in\mathcal{A}$. By The Set of Controls with Values in a Closed Bounded Set is Nonempty, Bounded, Convex and Closed, $\mathcal{U}_{\mathcal{A}}$ is nonempty, satisfies $\lVert\xi\rVert_{L^{2}}\le RT^{1/2}$ for every $\xi\in\mathcal{U}_{\mathcal{A}}$, is convex, and is closed for the topology determined by $d_{L^{2}}$. Claim numbers for $H$ refer to The Lebesgue Space of Square-Integrable Vector-Valued Functions is a Real Inner Product Space: the pairing is symmetric and linear in each argument, $\langle v,v\rangle_{L^{2}}=\lVert v\rVert^{2}_{L^{2}}$, the Cauchy-Schwarz inequality holds, and the norm vanishes only at $0$ (claims 4 and 5).

We also record two elementary facts about the natural powers of $2$ in $\mathbb{R}$, where $\iota$ denotes the canonical map of $\mathbb{R}$. First, every $2^{r}$ is positive, and $2^{s}\le2^{r}$ whenever $s\le r$. Indeed, by the recursion of the natural powers one has $2^{1}=2$ and $2^{r+1}=2\cdot2^{r}$, so positivity follows by Principle of Induction for the Natural Numbers, and then $2^{r}\le2\cdot2^{r}=2^{r+1}$ for every $r$; for $s\le r$ either $s=r$, or $r=s+i$ for some $i\in\mathbb{N}$ by claim 7 of Properties of the Order on the Natural Numbers, and induction on $i$ gives $2^{s}\le2^{s+i}$. Since all these numbers are positive, $2^{-r}\le2^{-s}$ for $s\le r$. Second, for every real $\varepsilon>0$ there is $J\in\mathbb{N}$ with $2^{-J}<\varepsilon$: by Principle of Induction for the Natural Numbers one has $\iota(J)\le2^{J}$ for every $J\in\mathbb{N}$ — for $J=1$ this reads $1\le2$, and if $\iota(J)\le2^{J}$ then $\iota(J+1)=\iota(J)+1\le2^{J}+2^{J}=2^{J+1}$, using $1\le2^{J}$ and claim 1 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field — hence $2^{-J}\le\iota(J)^{-1}$, and claim 3 of The Archimedean Property of the Real Numbers provides $J$ with $\iota(J)^{-1}<\varepsilon$.

Claim 1. Fix $\xi,\zeta\in H$ and put $a_{r}=\min\bigl(2^{-r},|\langle\xi-\zeta,w_{r}\rangle_{L^{2}}|\bigr)$. Each $a_{r}$ satisfies $0\le a_{r}\le2^{-r}\le2^{-1}$, so the set $\{a_{r}:r\in\mathbb{N}\}$ is nonempty and bounded above; its supremum exists because $\mathbb{R}$ is Dedekind complete. Thus $\rho(\xi,\zeta)$ is a well-defined nonnegative real number.

Symmetry. $\langle\zeta-\xi,w_{r}\rangle_{L^{2}}=-\langle\xi-\zeta,w_{r}\rangle_{L^{2}}$ by linearity, and the absolute value is unchanged, so the two defining sets coincide.

Vanishing on the diagonal. $\rho(\xi,\xi)=0$, since every $a_{r}$ is then $\min(2^{-r},0)=0$.

Triangle inequality. Let $\xi,\eta,\zeta\in H$ and $r\in\mathbb{N}$. Write $a=|\langle\xi-\eta,w_{r}\rangle_{L^{2}}|$, $b=|\langle\eta-\zeta,w_{r}\rangle_{L^{2}}|$ and $c=2^{-r}$; by linearity of the pairing and the triangle inequality for the absolute value, $|\langle\xi-\zeta,w_{r}\rangle_{L^{2}}|\le a+b$. We check $\min(c,a+b)\le\min(c,a)+\min(c,b)$: if $a+b\le c$ then $a\le c$ and $b\le c$, so the right-hand side is $a+b$; if $a+b>c$ and $a\ge c$ then the right-hand side is at least $c$; symmetrically if $b\ge c$; and if $a<c$ and $b<c$ the right-hand side is $a+b>c$. In all cases the inequality holds, and since $\min(c,a)\le\rho(\xi,\eta)$ and $\min(c,b)\le\rho(\eta,\zeta)$ we get $\min\bigl(c,|\langle\xi-\zeta,w_{r}\rangle_{L^{2}}|\bigr)\le\rho(\xi,\eta)+\rho(\eta,\zeta)$ for every $r$. The right-hand side is therefore an upper bound of the defining set, so $\rho(\xi,\zeta)\le\rho(\xi,\eta)+\rho(\eta,\zeta)$.

Separation. Suppose $\rho(\xi,\zeta)=0$ and put $v=\xi-\zeta$. For every $r$, $0\le\min(2^{-r},|\langle v,w_{r}\rangle_{L^{2}}|)\le0$, and $2^{-r}>0$, so $\langle v,w_{r}\rangle_{L^{2}}=0$. Let $\varepsilon>0$ be real. The set of terms of $(w_{r})_{r}$ is dense, so by Characterization of the Closure in a Metric Space by Open Balls there is $r$ with $\lVert v-w_{r}\rVert_{L^{2}}<\varepsilon$, whence by linearity and Cauchy-Schwarz

$$\lVert v\rVert^{2}_{L^{2}}=\langle v,v-w_{r}\rangle_{L^{2}}+\langle v,w_{r}\rangle_{L^{2}}=\langle v,v-w_{r}\rangle_{L^{2}}\le\lVert v\rVert_{L^{2}}\,\varepsilon .$$

If $\lVert v\rVert_{L^{2}}>0$, dividing by it gives $\lVert v\rVert_{L^{2}}\le\varepsilon$ for every real $\varepsilon>0$, which fails for $\varepsilon=\lVert v\rVert_{L^{2}}/2$. Hence $\lVert v\rVert_{L^{2}}=0$ and $\xi=\zeta$ by claim 5.

So $\rho$ is a metric on $H$, and its restriction to $\mathcal{U}_{\mathcal{A}}$ is a metric on $\mathcal{U}_{\mathcal{A}}$.

Claim 2. Suppose first that $\xi_{n}\rightharpoonup\xi$, and let $\varepsilon>0$ be real. Choose $J\in\mathbb{N}$ with $2^{-J}<\varepsilon$. For each $r$ with $r\le J$, weak convergence and linearity give that the real sequence $\bigl(\langle\xi_{n}-\xi,w_{r}\rangle_{L^{2}}\bigr)_{n}$ has limit $0$, so there is $N_{r}\in\mathbb{N}$ with $|\langle\xi_{n}-\xi,w_{r}\rangle_{L^{2}}|<\varepsilon$ for $n\ge N_{r}$. Let $N$ be the greatest element of the finite family $(N_{r})_{r\le J}$, which exists by Greatest Element of a Finite Family in a Totally Ordered Set. Let $n\ge N$ and $r\in\mathbb{N}$. If $r\le J$ then $\min\bigl(2^{-r},|\langle\xi_{n}-\xi,w_{r}\rangle_{L^{2}}|\bigr)\le|\langle\xi_{n}-\xi,w_{r}\rangle_{L^{2}}|<\varepsilon$; if $r>J$ then the minimum is at most $2^{-r}\le2^{-J}<\varepsilon$. So $\varepsilon$ is an upper bound of the defining set and $\rho(\xi_{n},\xi)\le\varepsilon$ for every $n\ge N$. Hence the real sequence $\bigl(\rho(\xi_{n},\xi)\bigr)_{n}$ has limit $0$.

Conversely suppose that sequence has limit $0$. Fix $r\in\mathbb{N}$. Since $2^{-r}>0$ there is $N$ with $\rho(\xi_{n},\xi)<2^{-r}$ for $n\ge N$; for such $n$ the minimum $\min\bigl(2^{-r},|\langle\xi_{n}-\xi,w_{r}\rangle_{L^{2}}|\bigr)$ is at most $\rho(\xi_{n},\xi)<2^{-r}$, so it cannot equal $2^{-r}$ and therefore equals $|\langle\xi_{n}-\xi,w_{r}\rangle_{L^{2}}|\le\rho(\xi_{n},\xi)$. Consequently, given a real $\varepsilon>0$ and choosing $N'\ge N$ with $\rho(\xi_{n},\xi)<\min(\varepsilon,2^{-r})$ for every $n\ge N'$, we obtain $|\langle\xi_{n}-\xi,w_{r}\rangle_{L^{2}}|\le\rho(\xi_{n},\xi)<\varepsilon$ for every $n\ge N'$. By linearity of the pairing the real sequence $\bigl(\langle\xi_{n},w_{r}\rangle_{L^{2}}\bigr)_{n}$ therefore has limit $\langle\xi,w_{r}\rangle_{L^{2}}$. Since $\lVert\xi_{n}\rVert_{L^{2}}\le RT^{1/2}$ for every $n$ and the set of terms of $(w_{r})_{r}$ is dense in $(H,d_{L^{2}})$, claim 5 of Basic Properties of Weak Convergence in the Lebesgue Space of Square-Integrable Vector-Valued Functions gives $\xi_{n}\rightharpoonup\xi$.

Claim 3. Let $(\xi_{n})_{n\in\mathbb{N}}$ be a sequence in $\mathcal{U}_{\mathcal{A}}$. Its terms satisfy $\lVert\xi_{n}\rVert_{L^{2}}\le RT^{1/2}$, so by Bounded Sequences in the Lebesgue Space of Square-Integrable Vector-Valued Functions Have Weakly Convergent Subsequences there are natural numbers $n_{1}<n_{2}<\dots$ and $\xi\in H$ with $\xi_{n_{j}}\rightharpoonup\xi$. The set $\mathcal{U}_{\mathcal{A}}$ is convex and closed, so $\xi\in\mathcal{U}_{\mathcal{A}}$ by Closed Convex Subsets of the Lebesgue Space of Square-Integrable Vector-Valued Functions are Weakly Sequentially Closed. By claim 2 the real sequence $\bigl(\rho(\xi_{n_{j}},\xi)\bigr)_{j}$ has limit $0$, that is, the subsequence $(\xi_{n_{j}})_{j}$ converges to $\xi$ in the metric space $(\mathcal{U}_{\mathcal{A}},\rho)$, and $\xi\in\mathcal{U}_{\mathcal{A}}$.

Thus every sequence in $\mathcal{U}_{\mathcal{A}}$ has a subsequence converging in $(\mathcal{U}_{\mathcal{A}},\rho)$ to a point of $\mathcal{U}_{\mathcal{A}}$, which is sequential compactness of $\mathcal{U}_{\mathcal{A}}$ as a subset of that metric space. By A Sequentially Compact Subset of a Metric Space is Compact it is therefore a compact subset of $(\mathcal{U}_{\mathcal{A}},\rho)$.