TITLE: Dual space of $H^1$
QUESTION [17 upvotes]: It holds that $W^{1,2}=H^1 \subset L^2 \subset H^{-1}$. This is clear since for every $v \in H^1(U)$, $u \mapsto (u,v)_{H^1}$ is an element of $H^{-1}$.
Moreover for every $v \in L^2(U)$, $u \mapsto (u,v)_{L^2}$ is an element of $H^{-1}$.
But I also know that $H^1$ is a Hilbert space and therefore it is isomorphic to its dual by the Riesz theorem.

My question is: how can there be $H^1(U) \subset L^2(U) \subset H^{-1}$ as well as $H^{-1}$ can be identified with $H^1(U)$?

REPLY [6 votes]: I'm essentially copying my answer to another closely related question. Note that the definition of $H^{-1}(\Omega)$ is different here.
Let me restate your question as follows. Let $\Omega$ be a nonempty open subset of $\mathbb{R}^d$. On the one hand, we have the sandwich:
$$
H^1(\Omega)\subset L^2(\Omega)\subsetneq H^{-1} (\Omega),\tag{*}
$$
where $H^{-1}(\Omega)$ is defined as  the dual of $H^1(\Omega)$. On the other hand, by the Riesz representation theorem, $H^{-1}(\Omega)$ can be identified as $H^1(\Omega)$. This suggests that we may write
$$
H^{-1}(\Omega)=H^1(\Omega)\tag{**}
$$
which contradicts $(*)$ since $L^2(\Omega)$ is a proper subspace of $H^{-1}(\Omega)$.
The problem is that $(**)$ is not true if one considers "$=$" in $(**)$ as equality of sets rather than an isomorphism of Hilbert space. 
As Terry Tao said:

"It is better to think of isomorphic pairs as being equivalent or identifiable rather than identical, as the latter can lead to some confusion if one treats too many of the equivalences as equalities. For instance, $\ell^2(\{0,1,2,\dots\})$ and $\ell^2(\{1,2,\dots\})$ are equivalent (one can simply shift the standard orthonormal basis for the former by one unit to obtain the latter), but one can also identify the latter space as a subspace of the former. It is fairly harmless to treat one of these equivalences as an equality, but of course one cannot do so for both equivalences at the same time."