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The following is the characterization theorem for $H^{-1}(U):=(H_0^1(U))^*$ in Evans's Partial Differential Equations:

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Here is my question:

The proof says that (iii) directly follows from (i). Would anybody elaborate why it is so?


  • $U$ is an open subset of $\mathbb{R}^{n}$
  • $H^{1}(U)$ is the Sobolev space of $L^{2}(U)$ functions with weak derivatives in $L^{2}(U)$
  • $H_{0}^{1}(U)$ is the closure of the subspace $\mathcal{C}_{c}^{\infty}(U)$ of compactly supported smooth functions on $U$
  • $H^{-1}(U)$ is the (continuous) dual of $H_{0}^{1}(U)$
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    $\begingroup$ I share your confusion. Isn't (iii) the definition of the inclusion? $\endgroup$ – Bananach Oct 5 '16 at 18:28
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    $\begingroup$ I do not like this part of the book of Evans. I prefer to characterize the $H^1$-norm through the Fourier transform and consequently we understand who is effectively the dual space. $\endgroup$ – user288972 Oct 5 '16 at 18:36
  • $\begingroup$ The Fourier characterization of Sobolev spaces only works for $U=\mathbb{R}^n$, so it's rather restrictive for a lot of applications. $\endgroup$ – Jeff Oct 5 '16 at 20:19
  • $\begingroup$ Yes, but after you just define $H^s_{0}(\Omega)=\overline{C_{c}^\infty(\Omega)}^{H^s(\mathbb{R}^n)}$, wtih $s \in \mathbb{R}$. Usually it goes like this, even without characterization of Fourier. Furthermore, it also can be shown that $H^s(\mathbb{R}^n)=H^m(\mathbb{R}^n)$, $s=m \in \mathbb{N}$, with equivalent norms. $\endgroup$ – user288972 Oct 5 '16 at 20:33
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    $\begingroup$ @JohnMartin Of course you can always define Sobolev spaces as the completion of $C^\infty$ under the Sobolev norms. Some books take this route, and others like Evans do not. Defining Sobolev spaces in this way is not, in and of itself, all that useful, since completion is an abstract construction. If you use the completion definition, you have to then prove that elements of the completion can be identified with functions, and that they have weak derivatives, etc., and you end up repeating what's in Evans' book anyway. $\endgroup$ – Jeff Oct 6 '16 at 0:18
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It's really a definition following from the inclusion $L^2(U) \subset H^{-1}(U)$. Perhaps it is better to write out the inclusion in more detail:

$$L^2(U) \sim (L^2(U))^* \subset (H^1_0(U))^* =: H^{-1}(U).$$

The inclusion above is true because the inclusion $H^1_0(U) \subset L^2(U)$ is continuous, i.e., $\|u\|_{L^2(U)} \leq \|u\|_{H^1_0(U)}$. So when we write $v \in L^2(U) \subset H^{-1}(U)$, what we really mean is that we are associating $v$ with the bounded linear functional on $L^2(U)$ given by

$$u \mapsto (v,u)_{L^2(U)} \ \text{for } u \in L^2(U),$$

which is also a bounded linear functional on $H^1_0(U)$ given by the restriction

$$u \mapsto (v,u)_{L^2(U)} \ \text{for } u \in H^1_0(U).$$

This is why we can write $\langle v,u\rangle = (v,u)_{L^2(U)}$ when $v \in L^2(U) \subset H^{-1}(U)$.

This is at least the canonical way to embed $L^2(U) \subset H^{-1}(U)$. We could, for instance, define $\langle v,u\rangle := 2(v,u)_{L^2(U)}$ for $v \in L^2(U)$ and $u \in H^1_0(U)$. This defines a bounded linear functional on $H^1_0(U)$, but is not canonical in the sense I described above.

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  • $\begingroup$ Thank you for your answer. Are you saying that (iii) should be taken as a definition instead of an assertion? $\endgroup$ – Jack Oct 6 '16 at 2:22
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    $\begingroup$ Yes, that's the way I interpret it. $\endgroup$ – Jeff Oct 6 '16 at 3:31

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