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Let $H^1(\Omega)$ be the Sobolev space $W^{1,2}$ defined over a domain $\Omega \subset \mathbb{R}^n$. It has been asked many times on this site, such as in this question, whether the space $H^{-1}$ is the dual of $H^1$ or $H^1_0$. It is usually said that $H^{-1}$ is the dual of $H^1_0$ as it can be shown that there exists an isomorphism between the two spaces, whereas the dual of $H^{1}$ cannot be characterized as easily (e.g. this question).

My confusion arises from the following theorem in the book Fourier Analysis and Nonlinear Partial Differential Equations by Bahouri, Chemin, and Danchin which describes the duality between $H^s$ and $H^{-s}$ for $s > 0$ and $s \in \mathbb{R}$:

For any real $s$, the bilinear functional $$\mathcal{B}:\begin{cases} \mathcal{S} \times \mathcal{S} \rightarrow \mathbb{C}\\(\phi, \varphi) \mapsto \int_{\mathbb{R}^d} \phi(x) \varphi(x) dx \end{cases}$$ can be extended to a continuous bilinear functional on $H^{-s} \times H^s$. Moreover if $L$ is a continuous linear functional on $H^s$, a unique tempered distribution $u$ exists in $H^{-s}$ such that $$\forall \phi \in \mathcal{S}, \langle L, \phi \rangle = \mathcal{B}(u, \phi).$$ In addition, we have $\|L\|_{(H^s)'} = \|u\|_{H^{-s}}$.

Here $\mathcal{S}$ denotes the Schwartz space of functions.

This theorem suggests that $H^{-1}$ is indeed the dual of $H^1$ and not $H^1_0$, which disagrees with the answer in the questions I linked above, so I am quite confused on which space $H^{-1}$ is dual to.

CBBAM
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    On $\Bbb{R}^d$, the spaces $H^1$ and $H^1_0$ are equal (recall that $C^{\infty}_c$ is contained in and dense in the former). – peek-a-boo Oct 09 '24 at 20:52
  • @peek-a-boo Thank you. I guess that implies $H^{-1}$ is indeed the dual space of $H^1$ when $\Omega = \mathbb{R}^n$ but not when $\Omega$ is a strict subset? – CBBAM Oct 09 '24 at 20:57
  • Is there any condition where they coincide other than when $\Omega = \mathbb{R}^n$? – CBBAM Oct 09 '24 at 21:04
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    I don’t know of any characterizations, but I guess in ‘most’ cases $H^1_0$ will be a strict subset of $H^1$, mainly because if you’re only taking limits of compactly supported functions, then the limit will vanish on the boundary of the open set, and a general $H^1$ function need not vanish on the boundary (i.e the trace operator is not zero). – peek-a-boo Oct 10 '24 at 07:28
  • @peek-a-boo Thank you. – CBBAM Oct 10 '24 at 23:48

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