Example of an $H^{-1}$ function that isn't $L^2$

Question

I'm going back over some PDE and Sobolev space theory, and the following is puzzling to me. Consider a nice domain $\Omega$ and the space $H^1_0(\Omega)$ of functions with $L^2$ first derivatives, vanishing at the boundary in the trace sense. Evans (section 5.9, page 299) then defines the dual of $H_0^1$, $H^{-1}$. He then makes sure to emphasize that we cannot identify $H_0^1$ with its dual. Why not? Isn't $H_0^1$ a Hilbert space? Is there an example of a distribution $f\in H^{-1}$ such that $f\notin L^2$, for instance?

Edit: in 1 dimension, we know that $u\in H^1_0$ has a continuous "version", so that for instance I think the Dirac delta distribution must be in $H^{-1}$. What about higher dimensions, where $H^1_0$ functions are not guaranteed to be continuous?

Answer 1

We can identify $H^1_0$ with $H^{-1}$ by Riesz Representation Theorem, because $H^1_0$ is Hilbert space. But, in this case, is not a "good" idea to do this identification.

You can consult Functional Analysis, Sobolev Spaces and Partial Diferential Equations, by Brezis, in the page 136, he makes a comment about it, and you will understand that choice of "not identify $H^1_0$ with $H^{-1}$" in Theorem of Characterization of $H^{-1}.$

Answer 2

For example, for $\Omega=\mathbb{R}$, the distribution $\delta_0$ defined by $\delta_0(\varphi)=\varphi(0)$ belongs to $H^{-1}(R)$(in fact $\delta\in H^s(R)$ for $s<-1/2$) but don't belong to $L^2(\mathbb{R})$.