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This may be a rather dense question, but I would nevertheless be grateful for some guidance.

The question has to do with the Sobolev spaces $H^m (\Omega)$ on an open bounded domain $\Omega$ of $\mathbb{R}^n$, where $m \geq 0$ is integer. These are Hilbert spaces; thus the Riesz representation theorem tells us that there is a one-to-one and onto correspondence between elements of $H^m (\Omega)$ and elements of its dual $H^{-m} (\Omega)$. Yet, we also have the inclusion property $H^{m} (\Omega) \subset L_2 (\Omega) \subset H^{-m} (\Omega)$ (e.g. Oden and Reddy, Intro. to the Mathematical Theory of Finite Elements, p. 108). How can both of these hold? In other words, how can there be a one-to-one and onto correspondence between $H^m (\Omega)$ and $H^{-m} (\Omega)$, and yet the former is a strict subset of the latter?

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That's a problem pretty much every one learning about Sobolev spaces seems to encounter sooner or later, see here. –  lvb Aug 3 '11 at 11:45
    
@user14178: A bijection is not the same thing as equality. The set of all even numbers is in bijection with the set of all numbers, but is also a proper subset. (This is one of the possible definitions for an infinite set.) –  Zhen Lin Aug 3 '11 at 11:58

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The dual space is the space of continuous linear functionals, and for a statement like $H^m \subset (H^m)^*$ to make sense one has to define a way to interpret functions as functionals, i.e. a linear embedding $H^m \to (H^m)^*$. There are two seemingly natural candidates: $$i_R(u)(v):=\left < u,v \right >_{H^m}$$ and $$i_D(u)(v):=\left < u,v \right >_{L^2}$$ The strictness of the inclusion translates to the question whether the embedding is an isomorphism, and it turns out that $i_R$ is an isomorphism according to the representation theorem, but $i_D$ is not! However, in accordance with distribution theory, $i_D$ is universally used to identify functions with functionals.

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Thanks for the pointers and the clear explanation. The apparent paradox is resolved by realizing that the inclusions are effected with the L_2 duality pairing and not the H^1, as you said. –  Mark Rashid Aug 4 '11 at 6:05

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