# Convolution of tempered distribution with $C_c^\infty$ function is $C_c^\infty$

Let $$u \in \mathscr{E}'(\mathbb{R}^n)$$ be a distribution with compact support, and let $$\psi \in C_c^\infty(\mathbb{R}^n)$$. I would like to show that $$u * \psi = v$$ as tempered distributions, where $$\langle u * \psi, \phi \rangle := \langle u, \psi * \tilde{\phi} \rangle,$$ with $$\tilde{\phi}(x) := \phi(-x)$$, and $$v(x) := \langle u, y \mapsto \psi(x - y) \rangle.$$

Observe that we can define the pairing between $$u * \psi$$ and an arbitrary distribution $$f \in \mathscr{S'}(\mathbb{R}^n)$$ as $$\langle u * \psi, f \rangle := \langle u, \left( x \mapsto \langle f, \psi(\cdot - x) \rangle \right) \rangle.$$ Using this definion, $$v(x) = \langle u * \psi, \delta_x \rangle,$$ where $$\delta_x$$ is the delta distribution at $$x$$. I then tried to show that for $$\phi \in \mathscr{S}(\mathbb{R}^n)$$, $$\langle u * \psi, \phi \rangle = \langle v, \phi \rangle,$$ by taking a mollifier $$(\chi_\varepsilon)_{\varepsilon > 0}$$, showing that $$\langle u * \psi, \chi_\varepsilon *\phi \rangle = \langle v, \chi_\varepsilon * \phi \rangle,$$ and then letting $$\varepsilon \to 0$$, but I have not managed to prove this.

Have a look at Proposition $$9.3$$ in Folland's Real analysis: Modern techniques and their applications (2nd edition). The idea is to approximate $$\langle v, \phi \rangle$$ with Riemann sums.