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.

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?

• It is just a direct application of the Riesz representation theorem. – Gustave Jun 8 at 23:03
• 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. – Sia Jun 8 at 23:13
• You must find the answer in this post math.stackexchange.com/questions/550619/… – 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.)

• In fact, formally, one is the Laplacian of the other. – 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 $$\langle e_m,e_{-n} \rangle = \frac{1}{\sqrt{1+4\pi^2 m^2}\sqrt{1+4\pi^2 n^2}} (e_m,e_{-n})$$)