Let $f:\mathbb{R}\to\mathbb{C}$ be a locally integrable function. For every $h>0$ we define $J_{h}f(x)=f\ast\frac{\chi_{[-h,0]}}{h}(x)$. I want to prove the following:
- $J_{h}f$ is continuous.
- If $f$ is a $C^{n}$ function then it is $J_{h}f$ and $(J_{h}f)^{(k)}=J_{h}f^{(k)}$.
But I've just been able to notice that $$J_{h}f(x)=\frac{1}{h}\int_{0}^{h}f(x+y)dy$$ and I can't see the continuity from here. I tried to use the density of the functions of compact support in $\mathscr{L}_{1}(\mathbb{R},\mathbb{C})$ but it seems to be not useful.
Any hint will be appreciated.
$\mathbb R$ is metric so we can use the sequential characterization of the continuity. Fix $x \in \mathbb R$ and take $(x_n)$ a sequence of real numbers converging to $x$. Then we use the continuity of the translations over $L^1$, indeed one can prove that whenever $g \in L^1(\mathbb R)$, $a \mapsto \tau_ag :=g(\cdot + a)$ is uniformly continuous from $\mathbb R$ to $L^1$. Now forall $n$, $$ |J_hf(x_n)-J_hf(x)| \leq \frac 1 h \int_0^h |\tau_{x_n}f - \tau_xf| $$ and noting $M = h+\max(|x|,|x_0|,...)$, $K=[-M,M]$, we have that $K$ is a compact set where $f$ is integrable. Take $\tilde f = f1_K \in L^1(\mathbb R)$, $$ \int_0^h |\tau_{x_n}f - \tau_xf| = \int_{x_n}^{x_n +h}|f-\tau_{x - x_n}f| \leq \int_{\mathbb R}|\tilde f - \tau_{x - x_n}\tilde f| = ||\tilde f - \tau_{x - x_n}\tilde f||_1 $$ and you are done.
Clearly you only need to see what happends for $n=1$. In this case you just need to differentiate under the integration symbol, so you only need to dominate $$ \left| \frac{d}{dx}f(x+y) \right| = |f'(x+y)|. $$ The classic trick is to fix $x_* \in \mathbb R$, and see that $$ \sup_{x \in [x_*-1,x_*+1]}\left| \frac{d}{dx}f(x+y) \right| \leq \sup_{x \in [x_*-1,x_*+1+h]}|f'(x)| $$ which is finite so integrable over $[0,h]$. Then you have the differentiability and the required formula at $x_*$ and thus everywhere.