This afternoon, while working out this answer by Nate Eldredge, I made some vain attempts at building cutoff functions of various kinds. Especially I was looking for the following.
Can a sequence $\zeta_n \in C^\infty(-1, 1)$ s.t.
- $0 \le \zeta_n \le 1$ and $\zeta_n(0)=1, \zeta(-1)=\zeta(1)=0$;
- $\lVert \zeta_n-1\rVert_2 \to 0$;
- $\lVert \zeta'_n \rVert_2$ is bounded
exist?
Graphically I was looking for something like this:
(in red the graph of $\zeta_n(x)=\exp(\frac{1/n}{x^2-1})$, in black the scaled graph of its first derivative). The task was to arrange things so that those black peaks stayed $L^2$-bounded.
After many unsuccessful trials I've come to the conclusion that such a $\zeta_n$ cannot exist. In fact, it should be $H^1$-bounded and so, up to a subsequence, $H^1$-weakly convergent. This implies $L^2$-weak convergence and so we should have $\zeta_n \stackrel{H^1}{\rightharpoonup}1$. But now we observe that $\zeta_n \in H^1_0(-1, 1)$. $H^1_0(-1, 1)$ is a norm-closed subspace of $H^1(-1,1)$, and so - because of its convexity - it is also weakly closed. We have thus gotten the contradiction $1 \in H^1_0(-1, 1)$.
Two questions:
- Is this reasoning correct?
- If 1. is affirmative, is there a more elementary way to see this? I've got the strong feeling of using a sledge-hammer to crack a nut.
Thank you for your attention.