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Let $H^s(\mathbb{R}^d):= \{ u \in \mathcal{S}' : (1+|\xi|^2)^{s/2}\hat{u}(\xi) \in L^2(\mathbb{R}^d)\}$. It can be shown that this space is a Hilbert space and that $H^s \subset H^t$ if $t \leq s$.

Now suppose we have $t > s>0$ such that $H^s \subsetneq H^t$. Then we know that $(H^s)^* \supsetneq (H^t)^*$, where $^*$ denotes the dual.

At the same time, since $H^s$ is a Hilbert space, the Riesz Representation theorem tells us that $H^s \cong (H^s)^*$ and similarly, $H^t \cong (H^t)^*$.

But then doesn't $(H^s)^* \supsetneq (H^t)^*$ imply that $H^s \supsetneq H^t$, which contradicts our earlier statement that $H^s \subsetneq H^t$?

What part of the argument here falls apart?

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When you say that $H^s \subsetneq H^t$, what you're saying is that there is a distinguished compatible family of strict inclusions $f_{s,t} : H^s \subsetneq H^t$. Dualizing these inclusions gives maps $f_{s,t}^{\ast} : (H^t)^{\ast} \to (H^s)^{\ast}$.

These maps don't appear to be inclusions to me. At least, one can't conclude this from abstract properties of Banach spaces; judging by the finite-dimensional case I would guess that the dual of an inclusion tends to be a quotient. So, if you like, the argument fails here.

But I think there is a second point you're missing, so I'll press on. All Hilbert spaces are canonically anti-isomorphic to their duals $d : H \to H^{\ast}$ (anti-isomorphic means that scalar multiplication is sent to its conjugate), hence in particular there are anti-isomorphisms $d : H^s \to (H^s)^{\ast}$ and $H^t \to (H^t)^{\ast}$. (You can upgrade these to isomorphisms by composing with the canonical anti-isomorphism between a Hilbert space and its "opposite" Hilbert space, but I don't think you get anything from doing this, especially since all it tells you is that the "opposite" and dual are naturally isomorphic.)

There's no possible way to get a contradiction from an argument like this because the maps $f_{s,t}$ are completely different from the maps $d$. (This is what happens when mathematical notation obscures what's really important, which is the morphisms, not the objects.) Hilbert spaces which strictly include each other can still be abstractly isomorphic; remember that all separable Hilbert spaces are isomorphic to $\ell^2$.


Willie brings up a point in the comments that is an extra source of confusion here: the Sobolev spaces all have different inner products, but in the first part of the argument I guess you were implicitly thinking of all of them as subspaces of $L^2$ with the induced inner product? Unfortunately the notation here also hides the choice of inner product. It really is quite awful.

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    $\begingroup$ Or, a cheaper way to think about it is: the implication $H^s \subset H^t \iff (H^s)^* \supset (H^t)^*$ requires taking the dual relative to (say) the $L^2$ pairing, whereas the isomorphism guaranteed by Riesz representation is given by the Hilbert-space norm pairing $H^s$ and $H^t$ respectively. $\endgroup$ – Willie Wong Mar 10 '11 at 23:38
  • $\begingroup$ @Willie: ah, I see; you're thinking of $H^s$ and $H^t$ as subspaces of $L^2$? $\endgroup$ – Qiaochu Yuan Mar 10 '11 at 23:40
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    $\begingroup$ The way I read the question is that the OP is interpreting $(H^s)^* = H^{-s}$, which is why you have the reversal of inclusions under duality. There is a natural inclusion of $H^s$ into $H^{-s}$, and that is different from the Hilbert space isomorphism of $H^s$ with $H^{-s}$. In essence, there are several maps involved in this discussion with overlapping domain and range, and one source of confusion is that one is tempted to think they are all the same map, when in fact they are vastly different. $\endgroup$ – Willie Wong Mar 11 '11 at 0:52
  • $\begingroup$ Like you say, if one keeps track of the inner product or exactly what each map is and how each map is defined, the confusion should be easily lifted. $\endgroup$ – Willie Wong Mar 11 '11 at 0:53
  • $\begingroup$ Yeah, I think I understand what you guys are saying, and I guess it is also why I was confused to begin with. $\endgroup$ – user1736 Mar 11 '11 at 3:17
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All the Sobolev spaces $H^s$ are isomorphic as Hilbert spaces simply because they have the same dimension.. the isomorphism from $H^s$ to $H^t$ is given by multiplying the Fourier transform by $(1 + |\xi|^2)^{s - t \over 2}$.

When one says that $H^s \subset H^t$ for $s > t$ one is looking at the finiteness of $||f||_{H_s} = ||(1 + |\xi|^2)^{s \over 2}\hat{f}(\xi)||_{L^2}$ and similarly for $H_t$. If you want a natural dual spaces that reverse the inclusion you have to view them as subspaces of a fixed Hilbert space with a single inner product, say that of $H^t$. In this case one can view the dual space of $H^s$ as $H^{-s}$ and the dual space of $H^t$ as $H^{-t}$ by taking $<f, g> = \int \hat{f}(\xi)\bar{\hat{g}(\xi)}\,d\xi$. In this situation the reversal of inclusions is clear.

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