# Existence of sequence of test functions converging to a constant

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 $$L^{2}$$.

• By your P.S. do you mean that the inequality should be $\lesssim$ (which is \lesssim in tex)? If so, this means that there exists a constant $C$ such that $\|\phi_\epsilon'\|_{L^2(0,1)} \le C$. Mar 12, 2023 at 23:27
• @RhysSteele Yes exactly, thank you for enlightening me Mar 13, 2023 at 7:31

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 $$\epsilon\to 0$$.
• Thank you very much, I've accepted your answer. I couldn't help but notice that this argument fails if we instead required $\|\phi_{\epsilon}'\|_{L^{1}} \le C$. Do you think the statement would still be false with this new hypothesis? Mar 13, 2023 at 7:30
• @duelspace with the $L^2$ norm replaced by the $L^1$ norm the statement is true. Let $\psi$ be a smooth non-negative function such that $\psi(x) = 0$ for $x \in [0,1/4]$ and $\psi(x) = 1$ for $x \in [3/4,1]$. Define $\phi_\epsilon(x) = \psi(\epsilon^{-1}x)$ for $x \in (0, \varepsilon]$, $\phi_\epsilon(x) = 1$ for $x \in [\epsilon, 1-\epsilon]$ and suppose that $\phi_\epsilon(x) = \phi_\epsilon(1-x)$ for all $x \in (0,1)$. Then $\|\phi_\epsilon'\|_{L^1(0,1)} = 2 \| \varepsilon^{-1} \psi'(\varepsilon^{-1} \cdot) \|_{L^1(0, \varepsilon)} \lesssim \|\psi'\|_{L^\infty}$. Mar 13, 2023 at 9:01
• @RhysSteele I see, so the $L^{1}$ norm is the only one which cancels this factor of $1/\epsilon$ appearing in front of the mollifier. Thanks! Mar 18, 2023 at 11:11