Is there a $C^m$ approximation $f_\epsilon$ of the Heaviside function such that

$$f_\epsilon(x) = f_1(x/\epsilon) = \begin{cases} 0 & \text{ if } x < 0 \\ 1 & \text{ if } x/\epsilon \ge 1 \end{cases}$$ $$\left|\frac{d^k}{dx^k}f_\epsilon\right| \le \frac{C}{\epsilon^k}$$ for $k \in \{1,2,\dots,m\}$ and some constant $C>0$, and $$\int\limits_{0}^{1} \left|\frac{d^k}{dx^k}f_\epsilon\right| dx \le \int\limits_0^1 |f_\epsilon| dx$$



1 Answer 1


$\newcommand\ep\epsilon$The answer is no. Indeed, assume that $$\int_0^1|f''_\ep(x)|\,dx\le\int_0^1|f_\ep(x)|\,dx\tag{0}$$ for $\ep\in(0,1)$. Let $$M:=\max_{0\le x\le\ep}|f_\ep(x)|.$$ Then $M\ge f_\ep(\ep)=1$ and $M=|f_\ep(u)|$ for some $u\in[0,\ep]$. So, $$M=|f_\ep(u)|=\max_{0\le x\le1}|f_\ep(x)|.\tag{1}$$ By the mean value theorem, $$M=|f_\ep(u)|=|f_\ep(u)-f_\ep(0)|=u|f'_\ep(v)|\le\ep|f'_\ep(v)|$$ for some $v\in[0,u]$. So, $$\frac M\ep\le|f'_\ep(v)|\le\int_0^v|f''_\ep(x)|\,dx\le\int_0^1|f''_\ep(x)|\,dx \le\int_0^1|f_\ep(x)|\,dx\le M,$$ by (0) and (1); thus we have a contradiction. $\Box$

  • $\begingroup$ This seems strange:by a change of variables, doesn't one get that the integral of the derivative $\approx \epsilon$, while the $L^1$ norm of $f_\epsilon$ is $\approx 1$? $\endgroup$
    – Hiro
    Jan 25, 2021 at 22:34
  • $\begingroup$ I mean $\int_0^1 f''_\epsilon(x) dx= \int_0^\epsilon f''_\epsilon(x)dx = \int_0^\epsilon f_1''(x/\epsilon) dx = \epsilon \int_0^1 f''_1(y)dy \le \epsilon$? $\endgroup$
    – Hiro
    Jan 25, 2021 at 22:50
  • $\begingroup$ @Hiro : In your latter comment, what are $f_1$ and $f_\epsilon$? I think it would be the best if you just re-check every step of my answer. There is nothing strange there: as you yourself conjectured, $f_\epsilon^{(k)}$ must be on the order of $1/\epsilon^k$. So, the integral of $f_\epsilon^{(k)}$ over the interval $[0,\epsilon]$ must be on the order of $1/\epsilon^{k-1}$. $\endgroup$ Jan 25, 2021 at 23:10
  • $\begingroup$ I edited the question to adjust the notation. But the $1/\epsilon^k$ estimate using the max. should be quite rough. Isn't it possible to use the change of variables as in my comment above to get that the integral goes like $\epsilon$? $\endgroup$
    – Hiro
    Jan 25, 2021 at 23:13
  • $\begingroup$ @Hiro : In the third integral in your second comment, the factor $1/\epsilon^2$ is missing. So, after your subsequent insertion of the restriction "$=f_1(x/\epsilon)$", the negative answer becomes much more immediate. $\endgroup$ Jan 26, 2021 at 0:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.