How to show that a Schwartz distribution is in a Lebesgue or Sobolev space? It is known that all $L^p$ spaces (and, consequently, all $W^{s,p}$ spaces) can be embedded in the space of Schwartz distributions $\mathcal D '$. There is a problem, though: how do I check whether some given distribution $u \in \mathcal D '$ belongs to any of the Lebesgue or Sobolev spaces mentioned above? (There are problems requiring the student to show this, and I have no clue what technique(s) to use and how to approach them.)
 A: Although hard to answer without specific examples in mind, this is a good question. The way I would approach this kind of problems is as follows. Let $\rho$ be a compactly supported smooth function on $\mathbb{R}^d$ with integral equal to $1$. Pick your favorite contant $L>1$. For $r\in\mathbb{Z}$, define $\rho_r(x)=L^{-rd}\rho(L^{-r}x)$ and let $u_r=\rho_r\ast u$. Namely, this is the smooth function
$$
u_r(x)=\langle u(y),\rho_r(x-y)\rangle_y
$$
which is a better way of writing the distributional pairing while retaining the expressive power of the merely formal
$$
\int_{\mathbb{R}^d}u(y)\rho_r(x-y)\ d^dy\ .
$$
Note that $\lim_{r\rightarrow -\infty}u_r=u$ in $\mathcal{D}'$ with its correct topology, i.e., the strong topology.
The first thing to do is check that $u_r$ is in $L^p(\mathbb{R}^d)$.
Then I would look for a bound of the form
$$
||u_r-u_{r-1}||_{L^p}\le K L^{\epsilon r}
$$
for some constants $K\ge 0$ and $\epsilon>0$.
If the bound holds, then $u_r$, $r\rightarrow -\infty$, will converge in $L^p$ to $u$ seen as an element of that subspace of $\mathcal{D}'$.
A: As has been mentioned in a comment your question is very broad. One possible answer is to use the space of tempered distributions $\mathcal{S}'(\mathbb{R}^n) \subset \mathcal{D}'(\mathbb{R}^n)$, it consists of the linear functional continuous with respect to $\mathcal{S}(\mathbb{R}^n)$ which is the Schwartz space.  In other words it can be shown that the Fourier operator $\mathcal{F}: \mathcal{S}'(\mathbb{R}^n) \longrightarrow \mathcal{S}'(\mathbb{R}^n)$ is a continuous (in the sense of distributions) isomorphism, extension of Fourier operator $\mathcal{F}: L^1(\mathbb{R}^n) \longrightarrow L^\infty(\mathbb{R}^n)$, and using $\mathcal{F}^{-1}  : \mathcal{S}'(\mathbb{R}^n) \longrightarrow \mathcal{S}'(\mathbb{R}^n)$ "you can go back". This probably can notice slightly in the spaces of Hilbert-Sobolev $H^s(\mathbb{R}^n)$, for example it can be shown that $\mathcal{E}'(\mathbb{R}^n) \subset \bigcup_{s \in \mathbb{R}} H^s(\mathbb{R}^n)$, where $\mathcal{E}'(\mathbb{R}^n)$ is the space of distributions with compact support, dual space of regular functions. I think that this is possible technique.
