I ’m studying about obstacle problems at the moment and I came across with this statement:
Let $V$ be a closed linear subset of $W^{1,2}(\Omega)$ where $\Omega$ is a bounded open connected subset of $\mathbb R^n$ and $\psi$ some measurable function.Then the set $\mathcal K=\{ v \in V: v \le \psi \}$ is obviously closed.
I ’m having a hard time understanding this "obviously" here. I have the following Theorem as a reference but I don’t see the connection:
THEOREM:Let $1\le p \le \infty$ and suppose that $\{u_n\}_n$ converges toward $u$ in $L^p(\Omega)$. Then there exists a subsequence $\{u_{nk}\}_k$ which converges to $u$ a.e in $\Omega$; moreover, there exists $u^{\ast} \in L^p(\Omega)$ such that $\vert u_{nk} (x) \vert \le u^{\ast}(x)$ for a.a $x\in \Omega$, any $k \in \mathbb N$
Should I consider $u^{\ast} \le \psi$? And what about $\nabla u_n$?
I would really appreciate if somebody could make this clear to me. Thanks in advance
You don't have to worry about any derivatives at all. If $v_n \leq \psi$ and $v_n \to v$ in $W^{1,2}$ then there is a subsequence $v_{n_k}$ converging almost everyehere to $v$. Just take limit in $v_{n_k} \leq \psi$ to get $v \leq \psi$.