Is $(H_0^1,\|\cdot\|_{L^2})$ a closed subspace of $L^2$?

Question

Let $-\infty<a<b<\infty$ and $f\in L^2(a,b)$.

Suppose $(f_n)$ is a sequence in $H_0^1(a,b)$ such that $\|f_n-f\|_{L^2}\overset{n\to\infty}{\longrightarrow}0$.

Can we conclude that $f\in H_0^1(a,b)$?

Answer 1

Daniel Fischer gave a good reason already, but here is a concrete example: $(a,b)=(-1,1)$, $$f(x)= \begin{cases}\frac{n}{n-1}(1-x), \quad &x>1/n \\ nx, \quad &0\le x\le 1/n \\ 0, \quad &x<-1/n\end{cases}$$ The functions $f_n$ converge in $L^2$ norm to $$f(x)= \begin{cases}1-x, \quad &x>0 \\ 0, \quad &x\le 0 \end{cases}$$ which is not in $H^1_0(a,b)$.