I have a hard time trying to show that an integral converges uniformly under those conditions:
I have: $f\in C_{c}^{0}({\mathbb{R}})$ and $g\in C^{k}(\mathbb{R})$ and I want to show that this integral is well defined:
$$(f*g)(x)= \int_{-\infty}^{\infty} f(t)g(x-t) \,dt $$
Can someone give me a hint on how to show that this fonction is in $C^{k}$
Well-definition is obvious. Let $\mathrm{supp}(f) \subseteq [a, b]$ where $0< a < b < \infty$. Then, because $[a, b]$ is compact, for every $x \in \mathbb{R}$ there exist $M, C \in \mathbb{R}$ such that $$ \lvert f(t) \rvert \leq M \text{ and } \lvert g(x-t) \rvert \leq C $$ for all $t \in [a, b]$. So $$ \int^\infty_{-\infty} \lvert f(t)g(x-t) \rvert ~\mathrm{d}t = \int^b_a \lvert f(t)g(x-t) \rvert ~\mathrm{d}t \leq MC(b-a) < \infty. $$
For differentiability fix $x \in \mathbb{R}$, consider the difference quotient and the fundamental theorem of calculus applied to $j(s) := g(x-t+sh)$ (note $j'(s) = hg'(x-t+sh)$), where $x, t \in \mathbb{R}$ are fixed: $$ \left \lvert \frac{(f *g)(x+h) - (f*g)(h)}{h} - \int_{b}^a f(t)g'(x-t)~\mathrm{d}t \right \rvert \leq \int^b_a \lvert f(t) \rvert \left \lvert \frac{j(1)-j(0)}{h} - g'(x-t) \right \rvert~\mathrm{d}t \leq \int^b_a \lvert f(t) \rvert \left( \int^1_0 \lvert g'(x-t+sh) - g'(x-t)\rvert~\mathrm{d}s \right)~\mathrm{d}t $$ Now use uniform continuity (why?) to make $\lvert g'(x-t+sh) - g'(x-t)\rvert$ smaller than any $\varepsilon > 0$. So what is $(f*g)'$ then and how can we calculate $f''$?