If $V$ is closed in $H^m(\Omega),\,m\geq1,$ is it also closed in $L^2(\Omega)$? In a proof on variational boundary value problems in the book "Introductory Functional Analysis" by Daya Reddy, the author uses the fact that if $V$ is closed in $H^m(\Omega),\,m\geq1,$ then $V$ is also closed in $L^2(\Omega).$ I was trying to reason this out by considering a sequence ${u_n}\in V$ such that $\lim_{n\rightarrow\infty}\lVert u_n-u\rVert_{L^2(\Omega)}=0$ and a contradictory statement that $u\not\in V$ but I wasn't able to see how to connect this to the fact that $V$ is complete in $H^m(\Omega).$ I'd appreciate any help.
Here is the relevant section from the Functional Analysis book by Daya Reddy, Chapter 9, Theorem 2:



 A: As commented by paul garrett, this is not true. You can pick $m = 1$,
then $V = H^1(\Omega)$ is certainly closed in $H^1(\Omega)$, but not in $L^2(\Omega)$.
However, one can proof the following:

Let $V \subset H^m(\Omega)$ be bounded, closed and convex. Then $V$ is closed in $L^2(\Omega)$.

The proof is rather straightforward: Let $u_n \to u$ in $L^2(\Omega)$ with $u_n \in V$. Then, $\{u_n\}$ is bounded in $H^m(\Omega)$.
Since $H^m(\Omega)$ is a Hilbert space, a subsequence of $u_n$ (not relabeled) converges weakly to some $\tilde u \in H^m(\Omega)$. It is easily checked that $\tilde u = u$. Further, since $V$ is closed and convex, it is weakly closed in $H^m(\Omega)$ and this yields $u = \tilde u \in V$.
Finally, I would like to mention that you can neither drop the boundedness (see above) or the convexity:
For example, take $m = 1$, $\Omega = (0,1)$ and
$$V = \{ v \in H_0^1(0,1) : |v'| = 1 \text{ a.e. in } (0,1)\}.$$
Then, $V$ is closed and bounded in $H^1(0,1)$ but not in $L^2(0,1)$. If I am not mistaken, the closure in $L^2(0,1)$ should be precisely the closed convex hull of $V$, e.g.
$$\{v \in H_0^1(0,1) : |v'| \le 1 \text{ a.e. in } (0,1)\}.$$
