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It is well known that for a given bounded domain $\Omega$, the Sobolev space $W^{1,2}(\Omega)$ is a Hilbert space, which is the space given by $$ W^{1,2}(\Omega)=\{u\in L^2(\Omega):\nabla u\in L^2(\Omega)\} $$ under the norm $$ \|u\|_{W^{1,2}(\Omega)}=\|u\|_{L^2(\Omega)}+\|\nabla u\|_{L^2(\Omega)}. $$ Then by Riesz representation theorem, the dual of this space should be isomorphic to the space itself. But I have seen in PDE books, the dual of $W^{1,2}(\Omega)$ is a bigger space than $W^{1,2}(\Omega)$, which is also not isomorphic to $W^{1,2}(\Omega)$, if I understood correctly. I could not understand the reason.

Can someone please help me to understand the concept of it?

Thank you.

Mathguide
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  • It's not quite that the $W^{1,2}$ norm is the sum of the $L^2$ norm of the function and the $L^2$ norm of its derivative... but, rather, that the norm-squared is the sum of the squares of those two norms. Also, some typos, $W^{k,2}$ should be $W^{1,2}$, evidently. But, anyway, those typoze do not affect the sense of your question. :) – paul garrett Jan 19 '22 at 18:41
  • Thanks. I corrected the typos. I hope the norm you mentioned is equivalent to the above norm. If you have some explanation about the question, kindly explain. Thanks again. – Mathguide Jan 19 '22 at 18:44
  • I'm pretty sure that the sum-of-norms norm is not topologically equivalent to the square-root of sums of squares norm... Maybe check your source? More later... – paul garrett Jan 19 '22 at 18:47
  • @paulgarrett These norms are of course equivalent as the $\ell_1$ norm and the $\ell_2$ norm on $\mathbb R^2$ are. – MaoWao Jan 19 '22 at 19:41
  • @MaoWao, ah, hm, perhaps so! Still, then, the sum-norm does not literally give a Hilbert-space structure...? In particular, why use the sum-of-norms norm at all? – paul garrett Jan 19 '22 at 19:44
  • @MaoWao, ah, yes, indeed, the two are topologically equivalent, for the reason you say. :) :) – paul garrett Jan 19 '22 at 19:50
  • You have to be careful about what topology you're taking the dual with. This notion of duality comes from taking the dual pairing to be the (real) $L^2$ pairing, see e.g. https://math.stackexchange.com/questions/740355/dual-of-h1-0-h-1-or-h-01?rq=1 – cmk Jan 20 '22 at 02:50

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All infinite-dimensional Hilbert spaces are isometrically isomorphic and the dual of $W^{1,2}(\Omega)$ is a separable Banach space. Thus, it is also isomorphic to $L^2(\Omega)$, $W^{2,2}(\Omega)$...

However, these different identifications result in different duality mappings. Therefore, it is recommended that you do not identify the dual spaces of $W^{k,2}(\Omega)$ for $k > 0$ with itself. You should only identify $L^2(\Omega)^*$ with $L^2(\Omega)$. In this way, you get a continuous embedding $$ E \colon W^{1,2}(\Omega) \to L^2(\Omega) $$ and the adjoint mapping $E^* \colon L^2(\Omega) \to W^{1,2}(\Omega)^*$ (here, we used the identification of $L^2(\Omega)^*$) is continuous and injective. Thus, you can identify $L^2(\Omega)$ with a subspace of $W^{1,2}(\Omega)^*$. Hence, $W^{1,2}(\Omega)$ is smaller than $L^2(\Omega)$, which is (in this sense) smaller than $W^{1,2}(\Omega)^*$.

gerw
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    Of course not all separable Banach spaces are isometrically isomorphic, but all separable infinite-dimensional Hilbert spaces are isometrically isomorphic. – MaoWao Jan 20 '22 at 12:36
  • @MaoWao: Yes, of course... – gerw Jan 20 '22 at 15:38