If compact embedding $H^1_0(\Omega) \hookrightarrow H^{-1}(\Omega)$ is not surjective, how can $H^{-1}(\Omega)$ be the dual of $H^1_0(\Omega)$? Let $\Omega$ be bounded with Lipschitz boundary.


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*By Rellich compactness, $\iota: H^1_0(\Omega) \hookrightarrow L^2(\Omega)$ is compact embedding. It is also dense.

*By Riesz representation, $L^2(\Omega) \overset{\sim}{\longrightarrow}(L^2(\Omega))^*$, so $L^2(\Omega)$ is identified with its dual.

*By Hahn-Banach, the dual map $\iota^*: (L^2(\Omega))^* \hookrightarrow (H^1_0(\Omega))^* = H^{-1}(\Omega)$ is dense compact embedding.


Thus, $\kappa: H^1_0(\Omega) \hookrightarrow H^{-1}(\Omega)$ is dense compact embedding. It is that the compact map in infinite dimensional space can not be surjective.
Here is my confusion:
$H^{-1}(\Omega)$ is dual of $H^1_0(\Omega)$. Thus, there must be an isometric isomorphism that maps every $x \in H^1_0(\Omega)$ to a dual $x^* \in H^{-1}(\Omega)$. So, the embedding $\kappa$ of $H^1_0(\Omega)$ to its dual should be surjective.
Then, how can $H^1_0(\Omega)$ is compactly embedded to its dual?

There exists a compact embedding $H^1_0(\Omega)\to H^{-1}(\Omega)$.  There also exists an isometric isomorphism $H^1_0(\Omega)\to H^{-1}(\Omega)$.  There is nothing contradictory about this, because these are two different maps.  (If you view a map from a normed space to its dual as a bilinear form on the space, the first map corresponds to the $L^2$ inner product restricted to $H^1_0(\Omega)$, while the second map corresponds to the standard inner product on $H^1_0(\Omega)$ that makes it a Hilbert space.  These are two different bilinear forms on $H^1_0(\Omega)$, so they give two different maps.)
