$f,g\in C_c^k(\mathbb R) \Rightarrow f*g\in C_c^{2k}(\mathbb R)$ If $f,g\in C_c^k(\mathbb R)$ für $k\in\mathbb N$, is $f*g\in C_c^{2k}(\mathbb R)$. I think this is not true. How can I prove it?
 A: Let us develop the proof of the statement methodically:
Firstly, $$f \ast g \in C(\mathbb R)$$ for $f, g \in C_c(\mathbb R)$. This follows from the dominated convergence theorem (or the theorem of the continuity of parameter integrals, which is in fact a corollary of the dominated convergence theorem). Moreover, one has $f \ast g \in C_c(\mathbb R)$ due to the fact that $\operatorname{supp} f \ast g \subseteq \operatorname{supp} f + \operatorname{supp} g$. In particular, the set on the right hand side is compact as $+ \colon \mathbb R \times \mathbb R \to \mathbb R$ is continuous and $\operatorname{supp} f$ and $\operatorname{supp} g$ are compact. Thus, it follows that also $\operatorname{supp} f \ast g$ is compact as a closed subset of a compact set.
Next it follows from the dominated convergence theorem (or the theorem of the differentiability of parameter integrals, which is another corollary of the dominated convergence theorem) that $$f \ast g \in C^1_c(\mathbb R) \qquad (f \in C^1_c(\mathbb R), \,g \in C_c(\mathbb R)).$$ This observation serves then as the base case of an induction over the order of regularity $k \in \mathbb N$. Thus, by induction, one obtains $$f \ast g \in C^k_c(\mathbb R) \qquad (f \in C^k_c(\mathbb R), \, g \in C_c(\mathbb R)).$$ Now there are (at least) two ways to deduce your desired statement. Either you prove with the dominated convergence theorem that
$$f \ast g \in C^{k + 1}_c(\mathbb R) \qquad (f \in C^k_c(\mathbb R),\, g \in C^1_c(\mathbb R))$$
and use induction over $\ell \in \mathbb N$ to obtain
$$f \ast g \in C^{k + \ell}_c(\mathbb R) \qquad (f \in C^k_c(\mathbb R),\, g \in C^\ell_c(\mathbb R))$$
or you use the argument below. Then the statement follows as the special case where $k = \ell$. However, one can use the another (more elegant) approach to prove the base case $\ell = 1$ and the induction steps. I just write it down for $\ell = 1$ since generalizing it is straightforward:
Let $f \in C^k_c(\mathbb R)$ and $g \in C^1_c(\mathbb R)$. Then integration by parts yields
\begin{align*}
f \ast g (x) &= \int_{\mathbb R} f(y) g(x - y) \, \mathrm d y \\
&= \biggl[F(y) g(x - y) \biggr]_{y = -\infty}^{y = \infty} - \int_{\mathbb R} F(y) g'(x - y) \, \mathrm d y = - F \ast g'(x)
\end{align*}
for all $x \in \mathbb R$, where $F \colon \mathbb R \to \mathbb R, \, F(x) := \int_{-\infty}^x f(y) \, \mathrm d y$. For each $h \in C^1_c(\mathbb R)$ one has that $\operatorname{supp} H, \operatorname{supp} h' \subseteq \operatorname{supp} h$ and since we know that $g' \in C_c(\mathbb R)$ and $F \in C_c^{k + 1}(\mathbb R)$ by the fundamental theorem of calculus, we can conclude that $f \ast g = -F \ast g' \in C_c^{k + 1}(\mathbb R)$ by what we have already shown.
The idea behind that argument is that you can shift regularity of functions in a convolution from one side to another by means of integration by parts and this is often very useful to get control over terms of poor regularity. In particular, if you have $f \in L^1_{\operatorname{loc}}(\mathbb R)$ (which is the bare mininum of regularity when you want all convolutions with functions of compact support to exist), then you have
$$f \ast g \in C^\infty_c(\mathbb R) \qquad (g \in C_c^\infty(\mathbb R)).$$
So convolutions with regular functions make even badly behaved functions regular. This idea of regularization is the back bone of the theory of distributions and the abstract reason that the heat equation is regularizing initial distributions in an abitrarily short time span $t > 0$ to smooth functions. This is namely just the result of the smoothness of the heat kernel. I hope things got a little bit more clear :-)
