Derivatives of convolution in $\mathbb{R}^n$

I am trying to prove that if $$f \in C_c^k(\mathbb{R}^n)$$ and $$g$$ is Lebesgue integrable on $$\mathbb{R}^n$$, then the derivatives of $$f * g$$ equal $$D^\alpha(f * g)(x) = \int_{\mathbb{R}^n}(D^\alpha f)(x-y)g(y)dy$$ and are continuous for all multi-indexes of order up to $$k$$.

I know the version for $$\mathbb{R}$$, however I am interested in this $$n$$-dimensional generalization.

Well, since $$f\in C_c^k$$, you have that $$D^{\alpha}f\in L^{\infty}$$, and thus, $$|D^{\alpha} f(x-y) g(y)|\leq \|D^{\alpha} f\|_{\infty} |g(y)|\in L^1$$. Thus, the desired simply follows from the dominated convergence theorem for differentiation under the integral, since for every fixed $$y$$,
$$D^{\alpha}( f(x-y)g(y))=D^{\alpha}(f)(x-y)g(y)$$
• Why is $D^\alpha f \in L^\infty$? I know that $f \in L^\infty$.
• $D^{\alpha} f$ is continuous, since $f\in C^k$ and there exists a compact set $K$ such that $f|_{\mathbb{R}^n\setminus K}\equiv 0$. Clearly, for $x\in \mathbb{R}^n\setminus K,$ we also have $D^{\alpha} f\equiv 0$ so $D^{\alpha} f\in C_c$. Any continuous function with compact support is $L^{\infty}$. Commented Jan 26, 2020 at 14:24