# First derivative of mollified functions

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.

• 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.. – 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 $c=0$.