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

I'm a bit confused, why do people not define $H^1(\Omega)^*$? Instead they only say that $H^{-1}(\Omega)$ is the dual of $H^1_0(\Omega).$

$H^1(\Omega)$ is a Hilbert space so it has a well-defined dual space. Can someone explain the issue with this?

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).
• So there is nothing wrong in me making use of $H^1 \subset L^2 \subset (H^1)^*$ as Hilbert triple, so if $f \in L^2$, the dual pairing $\langle f, u \rangle_{(H^1)^*, H^1} = (f,u)_{L^2}$ right? As your answer says this is true for $H^1_0$ and its dual but I see no reason why it can't work for $H^1$. – matt.w May 30 '13 at 10:35