Definition of the convolution with tempered distributions and Schwartz function 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
 A: Let us start by proving the inequality in the hint:
For fixed $x\in\mathbb{R}^{n}$, we have
$$
\partial_y^\alpha(\tau_x\tilde{\psi})(y) = \partial_y^\alpha(\psi(x-y)) = (-1)^{|\alpha|}(\partial^\alpha\psi)(x-y)
$$
and
$$
(1+|x-z|^2)^N \leq (1+2|x|^2+2|z|^2)^N \leq 2^N(1+|x|^2+|z|^2)^N \leq 2^N (1+|x|^2)^N(1+|z|^2)^N.
$$
Hence 
$$
\sup_{|\alpha|\leq N,y\in\mathbb{R}^{n}} (1+|y|^2)^{N} |\partial^\alpha (\tau_x\tilde{\psi})(y)| = \sup_{|\alpha|\leq N,z\in\mathbb{R}^{n}} (1+|x-z|^2)^{N}|\partial^\alpha \psi(z)|
\leq 2^N(1+|x|^2)^N q_N(\psi).
$$
Now observe that since $\tau_x\tilde{\psi}\in\mathcal{S}(\mathbb{R}^{n})$, for all $u\in\mathcal{S}'(\mathbb{R}^{n})$ and all $N\in\mathbb{N}$, there is $C_N>0$ such that
$$
|\psi*u(x)| = |\langle \tau_x\tilde{\psi},u\rangle |\leq C_n q_N(\tau_x\tilde{\psi}) \leq 2^N (1+|x|^2)^N C_N q_N(\psi).
$$
Since $u\ast\psi$ is continuous and therefore locally integrable, we get that $u\ast\psi$ is a distribution since, for $\varphi\in\mathcal{D}(\mathbb{R}^{n})$
$$
\langle \varphi, u\ast\psi\rangle = \int_{\mathbb{R}^{n}} \varphi(x)(u\ast\psi)(x)\;{\rm d}x.
$$
Using the above estimate for $N=1$, we get
$$
|\langle \varphi, u\ast\psi\rangle| \leq \int_{\mathbb{R}^{n}} |\varphi(x)||(u\ast\psi)(x)|\;{\rm d}x \leq 
2 C_1 \int_{\mathbb{R}^{n}} (1+|x|^2) \frac{(1+|x|^2)^{n+2}}{(1+|x|^2)^{n+2}} |\varphi(x)|\;{\rm d}x
\leq 2 C_1 q_1(\psi) q_{n+2}(\varphi) \int_{\mathbb{R}^{n}} \frac{{\rm d}x}{(1+|x|^2)^{n+1}} \leq C q_{n+2}(\varphi),
$$
where $C:=2 C_1 q_1(\psi)\int_{\mathbb{R}^{n}}\frac{{\rm d}x}{(1+|x|^2)^{n+1}}$.
Now the claim should follow from the fact that $\mathcal{D}(\mathbb{R}^{n})\subset\mathcal{S}(\mathbb{R}^{n})$ is dense.
A: I am also learning this stuff at the moment so I'll let you know what I think. I don't see how you would use closed graph theorem, in fact, you would need either $\mathcal{S}=\mathcal{S}(\mathbb{R}^d)$ to be a Banach space or compact Hausdorff. I do not believe either conditions are satisfied ($\mathcal{S}$ is normable or compact, at least I don't think so). Instead, it might just be easier to prove (a) directly. Say you want to prove (a) with respect to $x$. Just take a sequence $x_n$ in $\mathbb R^n$ that converges to $x$, say. Then 
$$\lim_{n\to \infty} \langle\tau_{x_n} \tilde{\varphi},u\rangle= \langle\lim_{n\to\to \infty}\tau_{x_n} \tilde{\varphi},u\rangle=\langle \tau_x \tilde{\varphi},u\rangle$$
where the last inequality holds once you prove $\tau_{x_n}\varphi\to \tau_{x}\varphi$ in $\mathcal{S}$, which shouldn't be too hard to prove. For continuity with respect to $u$, you want to prove that the function $u\mapsto u*\varphi(x)$ is continuous for each pair $(x,\varphi)$. This follows by definition of convergence of tempered distribution since $u_n\to u$ as tempered distribution implies that
$$\lim_{n\to \infty} \langle\tau_{x} \tilde{\varphi},u_n\rangle= \langle\tau_{x} \tilde{\varphi},u\rangle.$$
I'm too lazy to do (a) w.r.t $\varphi$. But I would imagine it's the same sort of argument.
For (b) and (c), there is a proof in Classical Fourier Analysis by Grafakos, and it's Theorem 2.3.20. in the 3rd edition. Unfortunately I don't completely understand his proof, especially the Riemann sum part.
