Proposition 1: Let $(X,d)$ a metric space and $(Y, \mathscr{A} , \mu)$ a measure space. Let $f : X \times Y \to \mathbb{R}$ such that

$\forall x \in X: f(x , \bullet) \in L^1 (\mu)$,
$\forall y \in Y : f(\bullet, y)$ is continuous,
$\forall x \in X$ exists an open neighborhood $V$ of $x$ and $\exists g \in L^1(\mu)$ such that $\forall (x^ \prime,y) \in V \times Y$: $|f(x,y)| \leq g(y)$.

Then $u: U \to \mathbb{R}$, $u(x) := \int f(x,\bullet) d \mu$ is continuous.

Proof: Let $x \in X$. Let $V$ an open neighborhood of $x$ and $g \in L^1(\mu)$ such that property 3 holds. Let $(x_n)_{n \in \mathbb{N}}$ a sequence in $X$ converging to $x$. Without loss of generality we may assume that the sequence takes values in $V$. Then for all $n \in \mathbb{N}$ and $y \in Y$: \begin{equation} |f(x_n,y) | \leq g (y). \end{equation} Therefore Lebesgue’s theorem of dominated convergence concludes. $\square$

Proposition 2: Let $m,k \in \mathbb{N}$, $U \subset \mathbb{R}^m$ open and $(Y, \mathscr{A} , \mu)$ a measure space. Let $f : U \times Y \to \mathbb{R}$ such that

$\forall x \in U: f(x , \bullet) \in L^1 (\mu)$,
$\forall y \in Y : f(\bullet, y) \in C^1 (U)$,
$\forall x \in U$ exists an open neighborhood $V$ of $x$ and $\exists g \in L^1(\mu)$ such that $\forall (x^ \prime,y) \in V \times Y$: $\|\nabla (f(\bullet,y))(x^\prime)\| \leq g(y)$.

Then $u: U \to \mathbb{R}$, $u(x) := \int f(x,\bullet) d \mu$ satisfies $u \in C^1 (U)$ and for all $j \in \{1,\dots,m \}$ and $x \in U$: \begin{equation} \partial_j u (x) = \int \partial_j (f(\bullet,y))(x) d \mu (y). \end{equation}

Proof: Let $x \in U$. Let $V$ an open neighborhood of $x$ and $g \in L^1(\mu)$ satisfying property 3. Without loss of generality we may assume that $V$ is convex. Let $(h_n)_{n \in \mathbb{N}}$ a sequence in $\mathbb{R} \setminus \{0\}$ converging to $0$. Without loss of generality we may assume that $\forall n \in \mathbb{N}$: $x+h_n e_j \in V$. Then for all $n \in \mathbb{N}$: \begin{equation} \frac{u(x+ h_n e_j) - u(x)}{h_n} = \int \frac{f(x +h_n e_j,y) - f(x,y)}{h_n} d\mu (y). \end{equation} Using a corollary to the mean value theorem we obtain for all $y \in Y$ and $n \in \mathbb{N}$: \begin{equation} \bigg|\frac{f(x +h_n e_j,y) - f(x,y)}{h_n}\bigg| \leq \sup_{ x^\prime \in V} \| \nabla (f(\bullet,y))(x^\prime)\| \leq g (y). \end{equation} Therefore Lebesgue’s theorem of dominated convergence can be applied to conclude that $u$ is partially differentiable at $x$ in the direction $j$ and that \begin{equation} \partial_j u(x) = \int \partial_j(f(\bullet,y))(x) d\mu (y). \end{equation} Proposition 1 concludes that $\partial_j u$ is continuous. $\square$

Proposition 3: Let $m,k \in \mathbb{N}$, $h \in C_c^k (\mathbb{R}^m)$ and $\Phi \in L^1_{\mathrm{loc}} (\mathbb{R}^m)$. Let $u: \mathbb{R}^m \to \mathbb{R}$, \begin{equation} u(x):= \int_{ \mathbb{R}^m} h(y) \Phi(x-y) dy. \end{equation} Then $u \in C^k(\mathbb{R}^m)$ and for all $\alpha \in \mathbb{N}^n$ with $|\alpha|\leq k$ and $x \in \mathbb{R}^m$: \begin{equation} D^\alpha u(x)= \int_{ \mathbb{R}^m} D^\alpha h(y) \Phi(x-y) dy. \end{equation} Proof: Assume $k=1$. The general statement follows by repeated application. By the translation and inversion invariance of the Lebesgue measure for all $x \in \mathbb{R}^m$: \begin{equation} u(x) = \int_{ \mathbb{R}^m} h(y) \Phi(x-y) dy = \int_{\mathbb{R}^n} h(x-y) \Phi (y) dy. \end{equation} Let $f : \mathbb{R}^m \times \mathbb{R}^m \to \mathbb{R}$ $f(x,y) := h(x-y) \Phi(y)$. Evidently $f$ satisfies assumption 1 and 2 of proposition 2. Let $x \in \mathbb{R}^n$. Let $R \in (0, \infty)$ such that the support of $h$ is contained in $B(0,R)$. Let $\varepsilon \in (0, \infty)$. Let $x^\prime \in B(x,\varepsilon)$. Then the support of $h(x^\prime - \bullet)$ is contained in $\bar B(x,R+\varepsilon)$. Since for $y \in \mathbb{R}^n$ with $\|x^\prime - y \|< R$: \begin{equation} \| y-x \| = \| y-x^\prime + x^\prime - x\| < R + \varepsilon. \end{equation} Let $\chi$ be the characteristic function of $\bar B(x, R+ \varepsilon)$. Then for all $y \in \mathbb{R}^n$: \begin{equation} \| (\nabla f (\bullet,y))(x^\prime)\| \leq \sup_{z \in \mathbb{R}^n} \|\nabla h (z)|| | \chi(y) \Phi(y)|. \end{equation} Therefore condition 3 of proposition 2 is fullfilled and so proposition 2 concludes the proof. $\square$