A differentiation under the integral sign. I know that

if $f\colon \mathbb{R}^n \to \mathbb{R}$ is integrable on any
  measurable, according to the usual $n$-dimensional Lebesgue measure
  $\mu_y$, and bounded subset of $\mathbb{R}^n$, and if $g \in C^k(\mathbb{R}^n)$ is compactly supported, then the
  function$$h:x\mapsto \int_{\mathbb{R}^n} f(x-y)g(y)\,d\mu_y$$belongs
  to $C^k(\mathbb{R}^n)$, and its partial derivatives of order
  $\leqslant k$ are given by
$$D^{\alpha} h(x) = \int_{\mathbb{R}^n} f(x-y)D^{\alpha} g(y)\,d\mu_y$$

whose proof, whose author I thank again, is found here.
I was wondering whether this result can extend to any $g \in C^k(\mathbb{R}^n)\cap L^{\infty}(\mathbb{R}^n)$ with $f\in L^1(\mathbb{R}^n)$, but I cannot adapt the linked proof because $\chi_{L-x_0}$ can no longer be used.
 A: Suppose that $f\in L^1$ and that $g$ is continuously differentiable with $g$ and all first order derivatives $g_1, g_2, g_3,\cdots g_n$ in $L^{\infty}$. Then
$$
              h_j(x) = \int f(x-y)g_j(y)dy = \int f(y)g_j(x-y)dy
$$
are continuous functions of $x$.
Fubini's Theorem allows you to interchange orders of integration:
\begin{align}
  \int_{x_{j,a}}^{x_{j,b}}\left(\int f(x-y)g_j(y)dy\right)dx_j
   & = \int_{x_j,a}^{x_j,b}\left(\int f(y)g_j(x-y)dy\right)dx_j \\
   & = \int f(y)\left(\int_{x_{j,a}}^{x_{j,b}}g_j(x-y)dx_j\right)dy \\
   & = \left.\int f(y)g(x-y)\right|_{x_j=x_{j,a}}^{x_j=x_{j,b}}dy \\
   & = \left.\int f(y)g(x-y)dy\right|_{x_j=x_{j,a}}^{x_j=x_{j,b}} \\
   & = \left.\int f(x-y)g(y)dy\right|_{x_j=x_{j,a}}^{x_j=x_{j,b}}
\end{align}
The integrand of the first integral in $x_j$ is continuous in $x_j$. Therefore, by the Fundamental Theorem of Calculus, the first integral is differentiable in $x_{j,b}$ and the derivative with respect to $x_{j,b}$ is equal to the integrand. That means that the last integral on the right must also be differentiable with respect to $x_{j,b}$, and one has the equality
$$
         \frac{\partial}{\partial x_j}\int f(x-y)g(y)dy = \int f(x-y)g_j(y)dy.
$$
The expression on the right is continuous in $y$. Therefore, $\int f(x-y)g(y)dy$ is continuously (jointly) differentiable with derivative
$$
        D \int f(x-y)g(y)dy = \int f(x-y)(Dg)(y)dy.
$$
This is easily extended to higher order derivatives by induction.
