Derivative of convolution is the convolution with a derivative I am trying to solve this exercise:
Let $\alpha$ be a multi-index. Show that $\partial^{\alpha}(u * v)=(\partial^{\alpha}u)*v$, where $u\in C_0^{k}(\mathbb{R}^n)$ and $v\in L^1_{loc}(\mathbb{R}^n)$.
Remark: I've tried to solve them by induction; for the case $n=1$ everything works, but I did not have success on the other cases.
 A: First of all it suffices to show it for $n=1$. Then $\partial^{\alpha+1}(u\star v) = \partial\left( \partial^{\alpha}(u\star v) \right)$ by by induction is equal to $\partial\left( \partial^{\alpha}u \star v \right)$ which is equal to $\partial\left( \partial^{\alpha}u \right) \star v$ by case $\alpha = 1$ as $\partial^{\alpha}u$ has still compact support, and you are done.
Even you don't want to say that it suffice to show it for $n=1$ or if this does not seem obvious to you, the induction works identically : you proceed by induction on $d = |\alpha| = \sum_{i=1}^n \alpha_i$ and you want to prove that for each $u\in C_0^{d}(\mathbb{R}^n)$ and $v\in L^1_{loc}(\mathbb{R}^n)$ one has $\partial^{\alpha}(u\star v) = \partial^{\alpha}u \star v$.
You have done the case $|\alpha| = 0$ or $1$. If $|\alpha| = d+1$ for $d\in\mathbf{N}$ then necessarily one $\alpha_i$ is in $\mathbf{N}^{*}$, and then $\alpha'$ definied by ${\alpha'}_j = \alpha_j$ except for $j = i$ where ${\alpha'}_i = \alpha_i - 1$ satisfies $|\alpha'| = d$ and $\partial^{\alpha} = \partial_i \partial^{\alpha'}$, and you say that $\partial^{\alpha} (u\star v) = \partial_i \partial^{\alpha'}(u\star v) $ which is equal by induction to $\partial_i (\partial^{\alpha'}u \star v) $ as $u \in C_0^{d}(\mathbb{R}^n)$ (as $u \in C_0^{d+1}(\mathbb{R}^n)$ by hypothesis), $\partial_i (\partial^{\alpha'}u \star v) $ is equal (by the case $|\alpha| = 1$ as $\partial^{\alpha'} u \in C_0^{1}(\mathbb{R}^n)$ as $u \in C_0^{d+1}(\mathbb{R}^n)$) to $\partial_i (\partial^{\alpha'}u) \star v = \partial^{\alpha}u \star v$.
