Is the set $\{ u \in H^1(\Omega) : 0 < a \leq u(x) \leq b\quad \text{a.e.}\}$ closed?

Question

Is the set $\{ u \in H^1(\Omega) : 0 < a \leq u(x) \leq b\quad \text{a.e.}\}$ closed?

$\Omega \subset \mathbb{R}^1$ is an interval. There is an embedding into $C^0(\Omega)$. But not sure if this helps.

Answer 1

It's closed for domains in any dimensions, in particular one does not need an embedding into $C^0$. Here is a stronger statement.

For any domain $\Omega\subset \mathbb R^n$, any $p\in [1,\infty]$, and any $a,b\in \mathbb R$, the set $$F=\{u\in L^p(\Omega):a\le u \le b \text{ a.e.}\}$$ is closed in the norm topology of $L^p$.

Proof. We have $u\in F$ if and only if $$a\int_{\Omega} \chi_E\le \int_{\Omega} u \chi_E \le b \int_{\Omega} \chi_E \tag{1}$$ for every set $E\subset \Omega$ of finite measure. Since the functional $u\mapsto \int_{\Omega} u \chi_E $ is continuous on $L^p$, the condition (1) defines a closed set. Intersection over all $E$ as above preserves the property of being closed. $\quad\Box$

The statement about $H^1$ follows, since any sequence converging in $H^1$ also converges in $L^2$. One can also say that $H^1$ norm induces stronger topology (i.e., with more open sets) than $L^2$ norm.