I remember that if $f,g \in \mathcal{S}(\mathbb{R}^n)$ , then it is well-defined
\begin{align*}
\displaystyle (f \ast g)(x)= \int_{\mathbb{R}^n} g(x-y)f(y)dy=\int_{\mathbb{R}^n} (\tau_x \widetilde{g})(y)f(y)dy
\end{align*}
where $\widetilde{g}(x)=g(-x)$ e $(\tau_a g)(x)=g(x-a)$, obviously $\tau_x \widetilde{\varphi} \in \mathcal{S}(\mathbb{R}^n)$, and if $u \in \mathcal{S}'(\mathbb{R}^n)$, $\varphi \in \mathcal{S}(\mathbb{R}^n)$, define
\begin{align*}
\displaystyle (u \ast \varphi)(x)=\langle \varphi(x-y), u(y) \rangle = \langle (\tau_x \widetilde{\varphi})(y), u(y) \rangle , \forall \varphi \in \mathcal{S}(\mathbb{R}^n)
\end{align*}
The problem is to prove that if $\psi \in \mathcal{S}(\mathbb{R}^n)$ or $\psi \in \mathcal{D}(\mathbb{R}^n)$, and $u \in \mathcal{S}'(\mathbb{R}^n)$, then $\displaystyle \langle \psi , u \ast \varphi \rangle = \langle \psi \ast \widetilde{\varphi} , u \rangle$.
Proof. we assume initially $u \ast \varphi, \psi, \varphi \in \mathcal{S}(\mathbb{R}^n)$, in particular $u \ast \varphi \in \mathcal{S}(\mathbb{R}^n) \subset L^1(\mathbb{R}^n)$ and by definition $\langle \psi , u \ast \varphi \rangle$ it defines the distribution determined by $u \ast \varphi$ $\forall \psi \in \mathcal{D}(\mathbb{R}^n)$ or $\forall \psi \in \mathcal{S}(\mathbb{R}^n)$:
\begin{align*}
\langle \psi , u \ast \varphi \rangle &= \int_{\mathbb{R}^n} (\varphi \ast u)(x) \psi(x)dx \\
&=\int_{\mathbb{R}^n} \int_{\mathbb{R}^n} \varphi(x-y)u(y)dy \psi(x) dx \\
&= \int_{\mathbb{R}^n} u(y) \int_{\mathbb{R}^n} \varphi(x-y) \psi(x) dx dy \\
&= \int u(y) dy \int_{\mathbb{R}^n} \widetilde{\varphi}(y-x)\psi(x) dx \\
&= \langle \widetilde{\varphi} \ast \psi , u \rangle
\end{align*}
Now if $u \in \mathcal{S}'(\mathbb{R}^n)$ is fixed, we consider the inclusions $\mathcal{D}(\mathbb{R}^n) \subset \mathcal{S}(\mathbb{R}^n) \subset \mathcal{S}'(\mathbb{R}^n) \subset \mathcal{D}'(\mathbb{R}^n)$. Since $\mathcal{D}(\mathbb{R}^n)$ is dense $\mathcal{D}'(\mathbb{R}^n)$, it is dense also in $\mathcal{S}'(\mathbb{R}^n)$, i.e. if $\lbrace \varphi_\epsilon \rbrace_{\epsilon >0}$ is standard mollifier, then $u \ast \varphi_\epsilon \rightarrow u$ in $\mathcal{S}'(\mathbb{R}^n)$
and that is the limit (i) $\lim_{\epsilon \rightarrow 0^+} \int_{\mathbb{R}^n} (u \ast \varphi_\epsilon) \psi(x) dx =\langle \psi, u \rangle$ $\forall \psi \in \mathcal{D}(\mathbb{R}^n)$, Therefore we can apply the previous case. Similarly it is proved that the thesis is true $\forall \psi \in \mathcal{S}(\mathbb{R}^n)$ and $u \in \mathcal{S}'(\mathbb{R}^n)$, since (i) worth for the approximation theorem in Lebesgue spaces, in other words $\mathcal{S}(\mathbb{R}^n)$ is dense in $\mathcal{S}'(\mathbb{R}^n)$.
I would be grateful if you are confirmed that this proof is correct
Best Answer
In reference to the comments, that is, of the various topologies. It is known that the Schwartz class $\mathcal{S}(\mathbb{R}^n)$ is a Fréchet space and also that the space of test functions $\mathcal{D}(\mathbb{R}^n)$ is dense in $\mathcal{S}(\mathbb{R}^n)$. In particular $\mathcal{D}_K(\mathbb{R}^n) \hookrightarrow \mathcal{S}(\mathbb{R}^n) \hookrightarrow L^1(\mathbb{R}^n)$ are continuous inclusions. Let $\mathcal{S}'(\mathbb{R}^n)$ the topological dual space $\mathcal{S}(\mathbb{R}^n)$. Now 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
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. On $\mathcal{S}'(\mathbb{R}^n)$ we consider the topology $\sigma(\mathcal{S}'(\mathbb{R}^n), \mathcal{S}(\mathbb{R}^n))$, so that $u_k \rightarrow u$ in $\mathcal{S}'(\mathbb{R}^n)$ means that
In other words, according to this definition of tempered distributions, the limit in (i) applies in any case: if $\psi \in \mathcal{D}(\mathbb{R}^n)$ or if $\psi \in \mathcal{S}(\mathbb{R}^n)$, because they are precisely restrictions of distributions.
I think this should formalize the last part of the proof.