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What is the dual space of $ H¹(\Omega) = W^{1,2}(\Omega) $? What is the dual space of $ W^{m,p}(\Omega) $? I know for example that the dual space of $ L^{p}(\Omega) $ for $ 1 \le p < \infty $ is $ L^{q}(\Omega) $ where $ \dfrac{1}{p} + \dfrac{1}{q} = 1 $.

Guy Fsone
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user29999
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1 Answers1

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The dual space of any topological vector space $X$ is the space of all bounded linear forms on $X$. Hence, the dual space of $W^{m,p}(\Omega)$ is the space of all linear $F:W^{m,p}(\Omega)\to \mathbb{R}$ such that $|F(v)|\leq C\|v\|_{W^{m,p}}$ for all $v\in W^{m,p}(\Omega)$.

If you are looking for explicit characterizations:

  • In the Hilbert space setting $p=2$, you can of course identify $H^m(\Omega)$ with its dual using the Riesz representation theorem (see also this question).
  • For $p=2$, you can also construct Sobolev spaces $H^s$ for any real $s$ using Fourier transforms. Then you can show that the dual of $H^s$ is $H^{-s}$.
  • If $p>n$ for $\Omega\subset\mathbb{R}^n$, then $W^{1,p}(\Omega)$ embeds densely into the continuous functions, and hence its dual space contains the space of regular Borel measures (such as the Dirac delta).

For details, I recommend Adams, R.A. and Fournier, J.J.F, Sobolev Spaces, 2nd ed., Academic Press, 2003.

Community
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Christian Clason
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    It's not a good idea to think that the Riesz-Fischer (conjugate...) identification of every Hilbert space with its dual behaves well with respect to maps among Hilbert spaces and adjoints thereof. It is rarely compatible... In particular, it is wildly incompatible with the inclusion $H^s\rightarrow H^{-s}$ and such. – paul garrett Jul 17 '13 at 00:51
  • @paulgarrett: Since what you say is a bit too general, could you please illustrate it with some examples? – Alex M. Jan 24 '19 at 18:57
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    @AlexM., first, there is the specific point that (e.g., for $s>0$), ignoring the issue of complex conjugation (which is not the sticking point), Riesz's isomorphism of $H^s$ with its own dual is not the inclusion map $H^s\to H^{-s}$ (which has dense image, but is not surjective, and the image has strictly finer topology than the subspace topology). Also, generally, the Riesz isom is not "functorial": [cont'd] – paul garrett Jan 24 '19 at 19:02
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    ... [cont'd] for $T:V\to W$ a linear map of Hilbert spaces, and $f:V\to V^$ and $g:W\to W^$, it is rarely the case that the obvious square commutes. That is, $f=T^*\circ g \circ T$ only when $T$ is an isometry to its image (which is not the case for Sobolev spaces...) – paul garrett Jan 24 '19 at 19:04