In the book where I'm studying there is the following exercise.
If $x \in \mathbb{R}^n$, $\varphi \in \mathcal{S}(\mathbb{R}^n)$ and $u \in \mathcal{S}'(\mathbb{R}^n)$ we define $(u \ast \varphi)(x)=\langle \tau_x \widetilde{\varphi} , u \rangle$, where we place $(\tau_x \widetilde{\varphi})(y):=\widetilde{\varphi}(y-x):=\varphi(x-y)$. Then
(a) $(u \ast \varphi)(x)$ is continuous with respect to $u \in \mathcal{S}'(\mathbb{R}^n)$, with respect to $\varphi \in \mathcal{S}(\mathbb{R}^n)$, and with respect to $x \in \mathbb{R}^n$
(b) $u \ast \varphi$ is a tempered distribution.
(c) If $\psi \in \mathcal{D}(\mathbb{R}^n)$, we have \begin{align*} \langle \psi, u \ast \varphi \rangle = u \left ( \int_{\mathbb{R}^n} \psi(x) (\tau_x \widetilde{\varphi})(\cdot) dx \right ) \end{align*} and extend this identity to case $\psi \in \mathcal{S}(\mathbb{R}^n)$.
HINT: To prove (b), check that $|(u \ast \varphi)(x)| \leq C(1+|x|^2)^N$ by proving an estimate $q_N(\tau_x \varphi) \leq 2^N(1+|x|^2)^N q_N(\varphi)$.
Note that for me there are these definitions. Let $\mathcal{S}'(\mathbb{R}^n)$ the topological dual space of $\mathcal{S}(\mathbb{R}^n)$. We have that the mapping $u \in \mathcal{S}'(\mathbb{R}^n) \longmapsto v=u_{|\mathcal{D}(\mathbb{R}^n)} \in \mathcal{D}'(\mathbb{R}^n)$ is linear and one-to-one because convergence in $\mathcal{D}(\mathbb{R}^n)$ implies convergence in $\mathcal{S}(\mathbb{R}^n)$, and $u_{|\mathcal{D}(\mathbb{R}^n)} \in \mathcal{D}'(\mathbb{R}^n)$ determines uniquely $u \in \mathcal{S}'(\mathbb{R}^n)$. Then a distribution $v \in \mathcal{D}'(\mathbb{R}^n)$ is the restriction of an element $u \in \mathcal{S}'(\mathbb{R}^n)$ if and only if there exist $N \in \mathbb{N}$ and a constant $C_N>0$ such that
\begin{align*} |u(\varphi)| \leq C_N q_N(\varphi)=C_N \sup_{x \in \mathbb{R}^n; |\alpha| \leq N} (1+|x|^2)^N |D^\alpha \varphi(x)| , \forall \varphi \in \mathcal{D}(\mathbb{R}^n) \end{align*}
where $q_N(\varphi)$ are seminorm that make $\mathcal{S}(\mathbb{R}^n)$ a Fréchet space. The elements of $\mathcal{S}'(\mathbb{R}^n)$ or their restriction to $\mathcal{D}(\mathbb{R}^n)$ are called tempered distributions.
To prove (a). I thought can be done with an application of the closed graph theorem, proving that
$\tau_a \cdot : \varphi \in \mathcal{S}(\mathbb{R}^n) \longmapsto \tau_a(\varphi)=\varphi(x-a) \in \mathcal{S}(\mathbb{R}^n)$
$\widetilde{\varphi} \cdot : \varphi \in \mathcal{S}(\mathbb{R}^n) \longmapsto \widetilde{\varphi}=\varphi(-x) \in \mathcal{S}(\mathbb{R}^n)$
are continuous with respect to convergence in $\mathcal{S}(\mathbb{R}^n)$. Is it correct, no?
Do you have any idea to the point (b) and (c)?
Note that (b) and (c) I tried to show in a different way, as here Tempered distributions and convolution
Thanks for any help