I have some questions about inequalties in the $H^1$ sense.
Defn: Let $u \in H^1(\Omega)$ and $E \subset \overline{\Omega}$. $u \geq 0$ on $E$ in the $H^1(\Omega)$ if there exists a sequence $u_n$ of Lipschitz continuous with modulus $1$ (written in this book as $u_n \in H^{1,\infty}$ functions) such that $u_n(x) \geq 0$ for $x \in E$ and $\|u_n - u\|_{H^1(\Omega)} \rightarrow 0$
A comment then follows:
Note that the functions satisfying the above bound: $\{ u \in H^1(\Omega)|u \geq 0 \}$ is a closed convex set. As such, it suffices to choose a sequence $u_m \rightarrow u$ weakly in $H^1(\Omega)$ by Banach Saks theorem.
1. I don't see this, could someone elaborate? Banach Saks has to do with when p.w. boundedness implies uniform boundedness right?
Here comes a theorem that discusses relation between inequalities in the $H^1$ sense and inequalities in the a.e. sense:
Supposed $\Omega \in \mathbb{R}^n$ is bounded, $E \subset \overline{\Omega}$ and $u\in H^1(\Omega)$. Then:
- $u \geq 0$ on $E$ in $H^1(\Omega) \implies u \geq 0$ on E a.e.
- $u \geq 0$ on $\Omega$ a.e $\implies$ $u \geq 0$ on $E$ in $H^1(\Omega)$
The proof of the second claim begins:
Let $v_n \in H^{1,\infty}$ satisfy $v_n \rightarrow u$ in $H^1(\Omega)$ and in $\Omega$ pointwise a.e.
2. Aren't these contradictory statements? Doesn't an appropriately modified typewriter sequence (one with triangles instead of "keys") converge in $H^1$ but not pointwise a.e.?
The point of the comment is indeed to relax the condition that $\|u_n-u\|_{H^1}\to 0$ to just assuming that $u_n\rightharpoonup u$ in the $H^1$ sense since the set of functions will be weakly closed by Banach Saks, as I explained in my comment.