Dual space of $H^1(\Omega)$

I don't quite agree with Dirk. The dual of $H^1$ sure is well-defined, it is just that it is hard to be characterized like $(H^1_0)^* = H^{-1}$, which is people usually say.

The isomorphism between $H^1_0$ and $H^{-1}$ can interpreted in boundary value problem sense: The Dirichlet Laplacian can be defined as $$ -\Delta^{\mathrm{Dir}}: H^1_0(\Omega)\to H^{-1}(\Omega) $$ in the sense that $$ \langle-\Delta^{\mathrm{Dir}} u,v \rangle = \int_{\Omega} \nabla u\cdot \nabla v, \quad \text{for } \forall u,v\in H^1_0(\Omega), $$ which is to say, by Riesz's representation, that $$\forall f\in H^{-1}(\Omega), \quad f = -\Delta^{\mathrm{Dir}} u \quad\text{for some } \; u\in H^1_0(\Omega).\tag{1}$$

Mimicking above, we can define Neumann Laplacian: $$-\Delta^{\mathrm{Neu}}: H^1(\Omega) \to H^1(\Omega)^*,$$ in the sense that $$ \langle-\Delta^{\mathrm{Neu}} u,v \rangle = \int_{\Omega} \nabla u\cdot \nabla v, \quad \text{for } \forall u,v\in H^1 (\Omega), $$ by the compatibility condition of a Neumann boundary value problem, the Neumann Laplacian $-\Delta^{\mathrm{Neu}}$ is rather a bijection from $H^1(\Omega)/\mathbb{R} \to (H^1(\Omega)/\mathbb{R})^*$.

Only if we use the full $H^1$-inner product to define an operator that maps an $H^1$-function to its dual: $$ \langle\mathcal{A} u,v \rangle = \int_{\Omega} (\nabla u\cdot \nabla v+u\,v),\quad \text{for } \forall u,v\in H^1(\Omega), $$ where $\mathcal{A}: H^1(\Omega) \to H^1(\Omega)^*$. However we don't have a (1) kind like statement because the unspecified boundary value.

As far as I remember, one usually defines $H^{-1}(\Omega)$ to be the dual space of $H^1(\Omega)$. The reason for that is that one usually does not identify $H^1(\Omega)^*$ with $H^1(\Omega)$ (which would be possible) but instead works with a different representation. E.g. one works with the $L^2$-inner product as dual pairing between $H^{-1}(\Omega)$ and $H^1(\Omega)$ (in case the element in $H^{-1}(\Omega)$ is an $L^2$-function).