I have a quick question about mollified functions and their derivatives. Let $\phi_\delta$, $\delta > 0$, be the standard mollifier. Let $f$ be a differentiable function and $f_\delta = f \ast \phi_\delta$ for $\delta > 0$, be a mollification of it. Is it true that $f_\delta' \to f'$ as $\delta \to 0$? What happens in the case, say when $f$ has a finite number of points where it has no derivative. Can we conclude that, if $f$ has a derivative at $x = c$, then $f_\delta'(c) \to f'(c)$ as $\delta \to 0$? What about when $f$ is not differentiable at $c$? What does $f_\delta'(c)$ approach as $\delta \to 0$ here (if there's a limit at all).

As a quick example take $f(x) = 1_{[0,N]}(x)\{x\}$ for some large enough real number $N > 0$, where $\{x\} := x - \lfloor x \rfloor$ is the fractional-part function. Note the function $\{x\}$ has derivative $1$ everywhere except on $\mathbb{Z}$, where it has jumps. Now $f_\delta = f \ast \phi_\delta$ which is $$ f_\delta(y) = \int_{0}^N\{x\} \phi_\delta(y - x) dx $$ and so $f_\delta' = f \ast \phi_\delta'$ is given by $$ f_\delta'(y) = \int_0^N \{x\}\phi_\delta'(y - x) dx. $$ Is it then true that for $y \in [0,N]\setminus \mathbb{Z}$, $f_\delta'(y) \to 1$ as $\delta \to 0$? What then of an improper integral over $y$ such as $$ \int_k^{k+1}\int_0^N \{x\}\phi_\delta'(y - x) dx dy. $$

Also, in practice one may sometimes need to upper bound integrals, such as $$ I := \int_{a}^b f(x) \phi_\delta'(y-x)dx, $$ uniformly for every $\delta >0$. In particular one is interested in what happens as $\delta \to 0$. Is there a bound we can use for $\phi_\delta'$ as $\delta \to 0$, so that one could have something like $I \leq M \int_a^b |f(x)| dx$ uniformly on $\delta$? Thanks a lot.

  • $\begingroup$ This is what I think too but I wanted to be sure (see a proof for example). Since taking derivatives is essentially taking a limit, there may be a double-limit action going on when you take $\delta \to 0$ (and so there may have to be some justification for interchanging limits for instance). My background is a bit weak on analysis and I'm not confident I'm being precise.. $\endgroup$ – user152169 Jan 11 '17 at 23:16

Claim: If $f$ is differentiable at a point $c$, then $f_\delta'(c)\to f'(c)$ as $\delta\to 0$.

Proof: Write $f(x)=f(c)+f'(c)(x-c)+r(x)$ where $r(x)/(x-c) \to 0$ as $x\to c$. Convolution with a symmetric mollifier preserves the linear part: that is, $$ f_\delta(x) = f(c)+f'(c)(x-c)+ (\phi_\delta*r)(x) $$ Differentiate this, placing the derivative on $\phi_\delta$ in the convolution (this is possible for any locally integrable $r$) $$ f_\delta'(x) = f'(c)+ (\phi_\delta'*r)(x) $$ At $x=c$, estimate the convolution using the fact that $\int |\phi_\delta'|$ is $C/\delta$ with $C$ independent of $\delta$: $$ |f_\delta'(c) - f'(c)| \le \frac{C}{\delta} \sup_{|x-c|\le \delta} |r(x)| $$ As $\delta\to 0$, the right hand side tends to $0$ as claimed. $\quad\Box$

If $f'(c)$ does not exist, the limit $\lim_{\delta\to 0} f_\delta'(c)$ may also fail to exist. One positive result that comes to mind: if both one-sided derivatives exist at $c$, then $\lim_{\delta\to 0} f_\delta'(c)$ exists and is equal to their average. For example: $f(x)=|x|$ with $c=0$.


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.