Dual space of $H^1$

Solution 1:

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.

Solution 2:

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."