Is it true that $L^2$ is compactly embedded in $(W^{1,2}_{0})^{\ast}$? Is it true that $L^{2}(\mathbb R^{n})$ is compactly embedded in $(W^{1,2}_{0}(\mathbb R^{n}))^{\ast}$?
If so, how can I prove it?
Context
I've just started to study Functional Analysis. I tried to read some articles about embeddings in Sobolev spaces, but all I can find are the standard inequalities (like Sobolev inequality etc).
 A: No, not on $\mathbb R^n$. Take a smooth function $\phi$ with support contained in the unit ball. Consider the sequence $f_n(x) = \phi(x-3ne_1)$ where $e_1$ is a basis vector. This is a sequence of bumps going into infinity. Translation by $3ne_1$ does not change either  $L^2$ norm or $W^{-1,2}$  norm (I write $W^{-1,2}$ for $(W_0^{1,2})^*$ here). Since the supports are separated, for $n\ne m$ 
$$\|f_n-f_m\|_{W^{-1,2}}\ge \frac{\langle f_n-f_m, f_n\rangle}{\|f_n\|_{W^{1,2}}}
= \frac{\langle f_n , f_n\rangle}{\|f_n\|_{W^{1,2}}} = c
$$ 
where $c>0$ does not depend on $n$, due  to translation invariance. Thus, $(f_n)$ has no convergent subsequence in $W^{-1,2}$.
Remarks
The argument applies to other Sobolev-type spaces with translation-invariant norms. Pretty much nothing is compactly embedded on $\mathbb R^n$, unless you use a weight decaying at infinity.  
On domains where the Rellich-Kondrachov theorem applies, you get compact embedding by dualizing the embedding of $W_0^{1,2}$ into $L^2$. The adjoint of a compact operator is compact.
