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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 $$\int\limits_{0}^{1} \left|\frac{d^k}{dx^k}f_\epsilon\right| dx \le \int\limits_0^1 |f_\epsilon| dx$$

hold?

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$\newcommand\ep\epsilon$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. $\Box$

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  • $\begingroup$ 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 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 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 $1/\epsilon^{k-1}$. $\endgroup$ – Iosif Pinelis Jan 25 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 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. $\endgroup$ – Iosif Pinelis Jan 26 at 0:45

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