Let $\Omega$ be bounded with Lipschitz boundary.

  1. By Rellich compactness, $\iota: H^1_0(\Omega) \hookrightarrow L^2(\Omega)$ is compact embedding. It is also dense.
  2. By Riesz representation, $L^2(\Omega) \overset{\sim}{\longrightarrow}(L^2(\Omega))^*$, so $L^2(\Omega)$ is identified with its dual.
  3. 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?

  • $\begingroup$ It is just a direct application of the Riesz representation theorem. $\endgroup$ – Gustave Jun 8 at 23:03
  • $\begingroup$ That is true. My confusion is, if $\kappa$ is compact (so not surjective), there must be an $x^* \in H^{-1}(\Omega)$ that does not have a corresponding $x \in H^1_0(\Omega)$. If by Riesz, $H^{-1}(\Omega)$ and $H^1_0(\Omega)$ then the relation between to is an isomorphism, rather than compact embedding. $\endgroup$ – Sia Jun 8 at 23:13
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    $\begingroup$ You must find the answer in this post math.stackexchange.com/questions/550619/… $\endgroup$ – Gustave Jun 8 at 23:20

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.)

  • $\begingroup$ In fact, formally, one is the Laplacian of the other. $\endgroup$ – Nate Eldredge Jun 9 at 3:29

Why not take a concrete example

  • $L^2(0,1)$ 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 $\langle e_m,e_{-n} \rangle = \frac{1}{\sqrt{1+4\pi^2 m^2}\sqrt{1+4\pi^2 n^2}} (e_m,e_{-n})$)


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