Difference between $\mathcal{E^{\prime}}$ and $\mathcal{D^{\prime}}$ What's the difference between $\mathcal{E^{\prime}}$(the space of compactly supported distributions) and $\mathcal{D^{\prime}}$ (the space of smooth compactly supported distributions)? Examples would be appreciated. 
 A: The identification of functions with distributions uses the pairing 
$$ (f,\phi) = \int f\phi \, dx $$
Where $f$ is at least locally integrable, and $\phi$ is compactly supported smooth. We choose these $\phi$ as test functions so that (a) every locally integrable function defines a distribution in the sense of a continuous functional, and (b) by using the formula prototyped by integration by parts we can apply any differential operator to the distributions. Thus $\mathcal{D}'$ is the space of distributions, and is dual to compactly supported functions. On the other hand $\mathcal{E}'$ is dual to the space of all smooth functions, and is identified as a subspace of $\mathcal{D}'$ because
$$ C_0^\infty \subset C^\infty$$
Dualizes to 
$$ \mathcal{E}' \subset \mathcal{D}'$$
Also, a point. the space $\mathcal{D'}$ is dual to compactly supported smooth functions, but is just called distributions (not compactly supported smooth distributions as you wrote), as it includes all manner of horribly not regular functions and distributions. On the other hand $\mathcal{E}'$ includes elements that have compact support themselves as distributions. There's a distinction between the support of the functions we "feed" to the distribution, and the support of the distribution itself. In general, the larger the class of test functions we allow, the smaller the class of distributions. $\mathcal{D'}$ is thus in some ways the largest we could hope for as compactly supported smooth functions are basically the most restrictive class of tests.
EDIT: If you're comfortable with sheaves, $\mathcal{E}'$ isn't a sheaf, while $\mathcal{D'}$ is. From a support perspective this is obvious, as impositions of compact support almost never allow one to define a sheaf. On the other hand if we have open sets $U_1 \subset U_2$ there is a canonical inclusion map $C_0^\infty(U_1) \to C_0^\infty(U_2)$ of extension by zero, which dualizes to a canonical restriction $\mathcal{D}'(U_2) \to \mathcal{D}'(U_1)$ and you can check these make distributions into a sheafon $\mathbb{R}^n$.
