Definition of the convolution with tempered distributions and Schwartz function

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.