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It holds that $W^{1,2}=H^1 \subset L^2 \subset H^{-1}$. This is clear since for every $v \in H^1(U)$, $u \mapsto (u,v)_{H^1}$ is an element of $H^{-1}$. Moreover for every $v \in L^2(U)$, $u \mapsto (u,v)_{L^2}$ is an element of $H^{-1}$. But I also know that $H^1$ is a Hilbert space and therefore it is isomorphic to its dual by the Riesz theorem.

My question is: how can there be $H^1(U) \subset L^2(U) \subset H^{-1}$ as well as $H^{-1}$ can be identified with $H^1(U)$?

tba
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2 Answers2

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You should not identify $H^{-1}$ with $H^1$, it leads to nothing but confusion. These spaces are dual to each other, but we do not think of the duality map as the identity map. The elements of these spaces have different meanings to us: $H^1$ consists of reasonably nice functions, $H^{-1}$ has some ugly distributions among its elements. So it makes sense that $H^1$ should be a proper subset of $H^{-1}$.

For that matter, all separable Hilbert spaces are isomorphic to each other, but this does not mean we should think that they are the same space.

But it's usually fine to identify $L^2$ with its dual. This allows for the following: the adjoint of the inclusion map $\iota:H^1\to L^2$ is a linear operator $\iota^*:(L^2)^* \to (H^1)^*$, that is, $\iota^*:L^2\to H^{-1}$. So we can interpret two inclusions $H^1 \subset L^2 \subset H^{-1}$ as adjoints of each other.

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    Why is it fine for us to identify $L^2$ with its dual (using the canonical identification brought to us by the inner product), but not in the case of $H^1$? What is the fundamental difference (Other than: In this particular instance, it is very convenient to do so)? – Sam Feb 25 '13 at 18:26
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    @Sam Stepping back from the particular function-space interpretation, consider two Hilbert spaces $U, V$. We don't know anything about these spaces, other than there is an isometric embedding $U\to V$, which is not surjective. The adjoint of this map goes the opposite way: $ V^\to U^$, also not surjective. Now we can choose to think of $U^$ as identical to $U$, or of $V^$ as identical to $V$, but not both at the same time. At least not before breakfast. –  Feb 25 '13 at 18:38
  • @Sam I was having these doubts. I kind of understood by realising that the Riesz identification map $R:H^1 \to H^{-1}$ is not the same as the composition of $i^\circ i$ in 5pm's notation. I don't know if this helps you. So you can identify $H^1$ with its dual with no problems as long as you don't say that identification map is the same as $i^ \circ i$ (correct me if I'm wrong..). – Lemon Feb 25 '13 at 18:40
  • To second 5pm's comments: it is misguided to too-automatically identify a Hilbert space and its dual (with or without worrying about complex conjugation). Indeed, this identification is compatible with a continuous linear map $T:V\rightarrow W$ of Hilbert spaces and the adjoint $T^*$ only under very restrictive hypotheses, as is easily seen as an exercise. In fact, this is an example of a "natural" map that is not compatible with other natural maps, illustrating a need for some precision in that business, too. – paul garrett Feb 25 '13 at 19:06
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    @Lemon, tba: NB: In the comment I wrote "isometric embedding" when I should have written "embedding", i.e., a bounded linear map. –  Feb 25 '13 at 19:06
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I'm essentially copying my answer to another closely related question. Note that the definition of $H^{-1}(\Omega)$ is different here.

Let me restate your question as follows. Let $\Omega$ be a nonempty open subset of $\mathbb{R}^d$. On the one hand, we have the sandwich: $$ H^1(\Omega)\subset L^2(\Omega)\subsetneq H^{-1} (\Omega),\tag{*} $$ where $H^{-1}(\Omega)$ is defined as the dual of $H^1(\Omega)$. On the other hand, by the Riesz representation theorem, $H^{-1}(\Omega)$ can be identified as $H^1(\Omega)$. This suggests that we may write $$ H^{-1}(\Omega)=H^1(\Omega)\tag{**} $$ which contradicts $(*)$ since $L^2(\Omega)$ is a proper subspace of $H^{-1}(\Omega)$.

The problem is that $(**)$ is not true if one considers "$=$" in $(**)$ as equality of sets rather than an isomorphism of Hilbert space.

As Terry Tao said:

"It is better to think of isomorphic pairs as being equivalent or identifiable rather than identical, as the latter can lead to some confusion if one treats too many of the equivalences as equalities. For instance, $\ell^2(\{0,1,2,\dots\})$ and $\ell^2(\{1,2,\dots\})$ are equivalent (one can simply shift the standard orthonormal basis for the former by one unit to obtain the latter), but one can also identify the latter space as a subspace of the former. It is fairly harmless to treat one of these equivalences as an equality, but of course one cannot do so for both equivalences at the same time."

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