I read somewhere that it is possible to find a sequence of test functions $\phi_{\epsilon} \in C^{\infty}_{c}(0,1)$ such that
- $\phi_{\epsilon} \ge 0$,
- $\phi_{\epsilon} \to 1$ as $\epsilon \to 0$,
- $\|\phi_{\epsilon}'\|_{L^{2}(0,1)} \lesssim 1 $ .
I wish to know if this claim is actually true or not. I am doubting this mainly because of the final statement which says that the derivatives are uniformly bounded. The reason I doubt this is because if this sequence is converging to $1$ and each element has compact support then surely the derivatives would have to be 'really large' near the endpoints $x=0$ and $x=1$? Intuitively I don't see how you could cook up a sequence of functions which have compact support, converge to $1$ AND have derivative uniformly bounded in $L^{2}$.
Indeed this is false. Suppose there exists a constant $C$ such that $\|\phi'_\epsilon\|\leq C$ for all $\epsilon$ (all norms in this answer are the $L^2$ norm). Then for each $\delta>0$, $$|\phi_\epsilon(\delta)|=\left|\int_0^\delta\phi'_\epsilon\right|\leq\|\phi'_\epsilon\|\|1_{[0,\delta]}\|=C\sqrt{\delta}$$ by Cauchy-Schwarz. For $\delta$ sufficiently small relative to $C$, this means $\phi_\epsilon(\delta)$ is always close to $0$, and so cannot converge to $1$ as $\epsilon\to 0$.