It is well known that for a given bounded domain $\Omega$, the Sobolev space $W^{1,2}(\Omega)$ is a Hilbert space, which is the space given by $$ W^{1,2}(\Omega)=\{u\in L^2(\Omega):\nabla u\in L^2(\Omega)\} $$ under the norm $$ \|u\|_{W^{1,2}(\Omega)}=\|u\|_{L^2(\Omega)}+\|\nabla u\|_{L^2(\Omega)}. $$ Then by Riesz representation theorem, the dual of this space should be isomorphic to the space itself. But I have seen in PDE books, the dual of $W^{1,2}(\Omega)$ is a bigger space than $W^{1,2}(\Omega)$, which is also not isomorphic to $W^{1,2}(\Omega)$, if I understood correctly. I could not understand the reason.
Can someone please help me to understand the concept of it?
Thank you.
All infinite-dimensional Hilbert spaces are isometrically isomorphic and the dual of $W^{1,2}(\Omega)$ is a separable Banach space. Thus, it is also isomorphic to $L^2(\Omega)$, $W^{2,2}(\Omega)$...
However, these different identifications result in different duality mappings. Therefore, it is recommended that you do not identify the dual spaces of $W^{k,2}(\Omega)$ for $k > 0$ with itself. You should only identify $L^2(\Omega)^*$ with $L^2(\Omega)$. In this way, you get a continuous embedding $$ E \colon W^{1,2}(\Omega) \to L^2(\Omega) $$ and the adjoint mapping $E^* \colon L^2(\Omega) \to W^{1,2}(\Omega)^*$ (here, we used the identification of $L^2(\Omega)^*$) is continuous and injective. Thus, you can identify $L^2(\Omega)$ with a subspace of $W^{1,2}(\Omega)^*$. Hence, $W^{1,2}(\Omega)$ is smaller than $L^2(\Omega)$, which is (in this sense) smaller than $W^{1,2}(\Omega)^*$.