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 $L^{2}$.

  • 2
    $\begingroup$ 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 $\|\phi_\epsilon'\|_{L^2(0,1)} \le C$. $\endgroup$ Mar 12, 2023 at 23:27
  • $\begingroup$ @RhysSteele Yes exactly, thank you for enlightening me $\endgroup$
    – duelspace
    Mar 13, 2023 at 7:31

1 Answer 1


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 $\epsilon\to 0$.

  • $\begingroup$ 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? $\endgroup$
    – duelspace
    Mar 13, 2023 at 7:30
  • 3
    $\begingroup$ @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 $\|\phi_\epsilon'\|_{L^1(0,1)} = 2 \| \varepsilon^{-1} \psi'(\varepsilon^{-1} \cdot) \|_{L^1(0, \varepsilon)} \lesssim \|\psi'\|_{L^\infty}$. $\endgroup$ Mar 13, 2023 at 9:01
  • $\begingroup$ @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! $\endgroup$
    – duelspace
    Mar 18, 2023 at 11:11

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