TITLE: dual of $H^1_0$: $H^{-1}$ or $H_0^1$?
QUESTION [22 upvotes]: I have a problem related to dual of Sobolev space $H^1_0$. By definition, the dual of $H^1_0$ is $H^{-1}$, which contains $L^2$ as a subspace. However, from Riesz representation theorem, dual of a Hilbert space is itself (in the sense of isomorphism). It seems that this implies dual of $H^1_0$ is $H^1_0$ itself.
From the result, I think the latter argument is not right. But where's the mistake?
This problem may related to the question about Hilbert space and its dual
A paradox on Hilbert spaces and their duals
However I couldn't understand the answers well after reading several times...
Thanks so much!

Thanks a lot for all the kind solutions!
The following are my conclusions:

Dual space strongly depends on action we take. For example, in our case, if we take $H^1$ inner product, the dual space is $H_0^1$ itself; if taking $L^2$ inner product, the dual space is larger than $L^2$; and taking all possible dual actions, the dual is by definition $H^{-1}$.
As dual of one space, there are certainly identifications between these spaces. Notice that identification between infinite dimensional spaces can be bijective and ISOMETRIC! Indeed there is a simple example: sequence {1,2,3...} and {2,4,6...}. Just taking metric of the latter to be half.
Another basic example is simple elliptic equation in $H_0^1$:
$$
\int_\Omega \nabla u\cdot \nabla v+\int_\Omega uv=\langle f,v\rangle , \forall v\in H_0^1(\Omega)
$$
where $f\in H^{-1}$. This gives isometric identification of $H^{-1}$ and $H_0^1$.
Brezis's book reminds that we can't take two different inner products to give both Riesz representation at the same time. The example there is very instructive.

REPLY [6 votes]: Riesz representation theorem is a : "representation" theorem,
and not an identification theorem. In the example 3 above, $f = - \Delta u + u$ can be represented by $u$, but cannot be identified by $u$, since
$f$ is in $H^{-1}$ whereas $u$ is in $H^1_0$.