Belonging to a compact convex and weak-* closure On p. 1430, we define $X_0$ to be the set of $(v,u,q)\in\mathcal{C}^\infty(\mathbb R^n_x\times\mathbb R_t)$ such that:


*

*$\operatorname{supp}(v,u,q)\subseteq\Omega$;

*$(v,u,q)$ solves (6) in $\mathbb R^n_x\times\mathbb R_t$;

*$(v(x,t),u(x,t),q(x,t))\in\mathcal U$ for all $(x,t)\in\mathbb R_x\times\mathbb R_t$.


Here:


*

*$\mathcal U=\operatorname{int}(K^{co}\times[-1,1])$ (the interior of the cartesian product of the convex hull of $K$ with $[-1,1]$), where $K=\{(v,w)\in\mathbb R^n\times\mathcal S_0^n:u=v\otimes v-\frac1n|v|^2I_n,|v|=1\}$;

*$\Omega$ is a bounded open domain in $\mathbb R^n_x\times\mathbb T_t$;

*(6) is essentially the Euler equation, which in the variables $v,u,q$ reads $\partial_tv+\operatorname{div}u+\nabla q=0$.


We then proceed to define $X$ as the weak-* closure of $X_0$ in $L^\infty$, and state the following.
Theorem
The set $X$ with the topology of $L^\infty$ weak-* convergence is a nonempty compact metrizable space. Moreover, if $(v,u,q)\in X$ is such that $|v(x,t)|=1$ for almost every $(x,t)\in\Omega$, then $v$ and $p:=q-\frac1n|v|^2$ is a weak solution of (1) [which is Euler, ndR] in $\mathbb R^n_x\times\mathbb R_t$ such that $v(x,t)=0$ and $p(x,t)=0$ for all $(x,t)\in\mathbb R^n_x\times\mathbb R_t\smallsetminus\Omega$.
The proof goes like this.
Proof.
We previously proved $0\in\mathcal U$, so that $0\in X_0$ and thus $X\neq\varnothing$. It is closed in $L^\infty$ by definition, and it is bounded because, by condition 3, $X_0$ is, and weak-* closure preserves boundedness. It is then well-known that the closed unit ball of a the dual of a separable space is weak-* compact (Banach-Alaoglu) and weak-* metrizable (IIRC – this is where separability comes in), and $L^1$ is separable and $L^\infty$ is its dual, so $X$ is a nonempty compact metrizable space.
«Since $\overline{\mathcal U}$ is a compact convex set, any $(v,u,q)\in X$ [1] satisfies $\operatorname{supp}(v,u,q)\subseteq\overline\Omega$, [2] solves (6), and [3] takes values in $\overline{\mathcal U}$». By the last claim, $(v,u)(x,t)\in K^{co}$ a,e,m and therefore $(v,u)(x,t)\in K$ iff $|v|=1$, which concludes the proof by Lemma 2.1.      $\diamond$
The sentence in guillemets is where I'm having trouble.
To prove [1], we observe that, if $V\in X$ (I'll avoid the triple for ease of writing), then there is a sequence $X_0\ni V_k\overset\ast\rightharpoonup V$ in $L^\infty$, but then we can show $V_k|_{(\overline\Omega)^C}\to V|_{(\overline\Omega)^C}$ because $f\in L^1((\overline\Omega)^C$ is contained in the "global" $L^1$ by setting $\tilde f=f$ on $(\overline\Omega)^C$ and $\tilde f=0$ on $\overline\Omega$, but $V_k|_{(\overline\Omega)^C}\equiv0$ by definition of $X_0$, so $V|_{(\overline\Omega)^C}\equiv0$, meaning $\operatorname{supp}V\subseteq\overline\Omega$.
[2] can be proved by IBP and then the weak-* convergence.
But what about [3]? The "hearsay" section of this question comes from asking about this, but of course we cannot argue that, since $T:=\{V\in\overline{\mathcal U}\}$ is obviously strong-closed and convex, then it's weak-* closed and thus contains $X$ because it contains $X_0$. So how do we prove $X\subseteq T$?
 A: In these lecture notes, we find the following theorem.
Theorem
Let $K\subseteq\mathbb R^d$ be bounded. If $v_n\in K$ a.e. and $v_n\overset\ast\rightharpoonup v$ in $L^\infty$, then $v\in\overline{K^{co}}$ a.e.
Of course, assuming this result, if we take $K$ closed and convex, we will have $v\in K$ a.e., thus proving [3]. Indeed, $\overline{K^{co}}$ can be characterized as the intersection of all closed convex sets containing $K$(1), and this operation is obviously idempotent, so that $\overline{(\overline{K^{co}})^{co}}=\overline{K^{co}}$.
Let us then prove the theorem.
Proof.
Let $L$ be an affine function and $L\geq0$ on $K$. $v_n\in K$, so $L(v_n)\geq0$. Since $L$ is affine, we will have $L(x)=a\cdot x+b$ for $a,b\in\mathbb R^d$. So $L(v_n)=a\cdot v_n(x)+b$, which clearly converges weak-* to $a\cdot v(x)+b=L(v)(x)$. Then, $L(v)\geq0$ a.e.. This means that, for all $H$ half-spaces containing $K$, $v\in H$ a.e.. This implies $v\in\overline{K^{co}}$ a.e.. Indeed, Hahn-Banach says any $x\notin\overline{K^{co}}$ can be separated from $\overline{K^{co}}$ by a hyperplane, so that $\overline{K^{co}}\supseteq\bigcap\{H\text{ half-plane containing }\overline{K^{co}}=:\mathbb H$. However, since $\overline{K^{co}}$ is contained in all those planes, we have that $\overline{K^{co}}\subseteq\mathbb H$, meaning it coincides with it. Since the half-planes containing $K$ are the same as those containing $\overline{K^{co}}$, and $v$ is a.e. in all of those, we conclude the proof.
(1) For any $C$ convex and closed you have $K\subseteq C\implies K^{co}\subseteq C\implies K^{co}\subseteq C$, and viceversa $\overline{K^{co}}$ is a closed convex set containing $K$, closed because it's a closure, and convexity is trivial.
