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?

Why not take a concrete example


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*$L^2(0,1)$ is the inner product space $\langle f,g\rangle= \int_0^1 f(x)\overline{g(x)}dx$,
$H^1(0,1)$ is the inner product space $( f,g) =\langle f,g\rangle+\langle f',g'\rangle$

*$e_n = \frac{e^{2i \pi nx}}{\sqrt{1+4\pi^2 n^2}}$ is an orthonormal basis of $H^1(0,1)$. 

*The map $f \mapsto (g \mapsto (g,\overline{f}))$ is an isomorphism $H^1(0,1) \to H^1(0,1)^*$, it sends $e_n$ to $(.,e_{-n})$

*The map $f \mapsto \langle .,\overline{f}\rangle$ is dense compact $H^1(0,1) \to H^1(0,1)^*$, 
it sends $e_n$ to $(.,\frac{e_{-n}}{1+4\pi^2 n^2})$
(as $\langle e_m,e_{-n} \rangle = \frac{1}{\sqrt{1+4\pi^2 m^2}\sqrt{1+4\pi^2 n^2}} (e_m,e_{-n})$)
