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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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    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.2011-03-10
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    @Willie: ah, I see; you're thinking of $H^s$ and $H^t$ as subspaces of $L^2$?2011-03-10
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    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.2011-03-11
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    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.2011-03-11
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    Yeah, I think I understand what you guys are saying, and I guess it is also why I was confused to begin with.2011-03-11
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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 $ = \int \hat{f}(\xi)\bar{\hat{g}(\xi)}\,d\xi$. In this situation the reversal of inclusions is clear.