# A special approximation of the Heaviside function

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?

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

• 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 $\approx 1$?
– Hiro
Jan 25 '21 at 22:34
• I mean $\int_0^1 f''_\epsilon(x) dx= \int_0^\epsilon f''_\epsilon(x)dx = \int_0^\epsilon f_1''(x/\epsilon) dx = \epsilon \int_0^1 f''_1(y)dy \le \epsilon$?
– Hiro
Jan 25 '21 at 22:50
• @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 $1/\epsilon^{k-1}$. Jan 25 '21 at 23:10
• 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 $\epsilon$?
– Hiro
Jan 25 '21 at 23:13
• @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. Jan 26 '21 at 0:45