Let $m, n$ be positive integers. Let $\Omega$ be an open subset of $\mathbb{R}^n$. Let $\mathcal{D}(\Omega)$ be the space of smooth functions with compact support in $\Omega$.
Now, let $A: \mathbb{R}^m \to \mathcal{D}(\Omega)$ be a test-function-valued map that satisfies the following three conditions:
- There exists a compact subset $L$ in $\Omega$ such that $ A(t) \in \mathcal{D}(L) $ for all $t \in \mathbb{R}^m $.
- The support of $A$ is locally uniformly compact, i.e. for any $x_0 \in \Omega$, there exists a neighbourhood $ U(x_0) \subseteq \Omega $ and a compact set $ K(x_0) \subseteq \mathbb{R}^m $ such that for any $x \in U(x_0)$ and $t \in \mathbb{R}^m \setminus K(x_0)$, we have $A(t)(x) = 0$.
- $A$ is uniformly continuous as a map into $\mathcal{D}(L)$, i.e. for any multi-index $\alpha$ and any $\epsilon > 0$, there exists $\delta > 0$ such that whenever $|s-t| < \delta$, we have $ ||\partial^\alpha A(s) - \partial^\alpha A(t)||_\infty < \epsilon $ on $L$.
The assertion is that $A$ has a (weak) vector-valued integral in $\mathcal{D}(\Omega)$.
More specifically, if we define the function
$$ \int A : \Omega \to \mathbb{R}, \quad (\int A)(x) := \int_{\mathbb{R}^m} A(t)(x) dt $$
then:
- $ \int A \in \mathcal{D}(\Omega) $, and
- for any distribution $u \in \mathcal{D}'(\Omega)$, we have $$ \langle u, \int A \rangle = \int_{\mathbb{R}^m} \langle u, A(t) \rangle dt $$
Question:
How to prove the assertion?