TITLE: difference between the dual space of $H^1(\Omega)$ and the dual of $H^1_0(\Omega)$
QUESTION [5 upvotes]: In the Partial Differential Equations by Evans (2nd edition p299), $H^{-1}(\Omega)$ denotes the dual space to $H^1_0(\Omega)$ where $\Omega$ is an open subset of $\mathbb{R}^n$ and $H^1(\Omega)=W^{1,2}(\Omega)$, $H^1_0(\Omega)=W^{1,2}_0(\Omega)$:
$$
W^{1,2}_0(\Omega)=\overline{C_c^\infty(\Omega)}^{\|\cdot\|_{W^{1,2}(\Omega)}}
$$
While in the Navier Stokes Equations by Constantin and Foias (p7), $H^{-1}(\Omega)$ denotes the dual space of $H^1(\Omega)$.  
Let $X$ be the (continuous) dual of $H^1(\Omega)$ and $Y$ the dual of  $H^1_0(\Omega)$. One has that $X\subset Y$. 
Here is my question: 

Could somebody describe the difference between $X$ and $Y$?

REPLY [3 votes]: Every element of $W^{m,p}(\Omega)'$ is the continuous extension of a distribution. However, the extensions are non-unique. By restricting oneself to $W^{m,p}_0(\Omega)$ the extensions are unique. This provides a characterization of the dual of $W^{m,p}_0(\Omega)$ as the space of all distributions $T \in D'(\Omega)$ of the form
$$T=\sum_{0\leq|\alpha|\leq m}(-1)^{|\alpha|}D^\alpha T_{v_\alpha}$$
where $v_\alpha \in L^{p'}$ and $T_{v_\alpha}(\phi)=\langle \phi,v_\alpha \rangle$, the duality pairing.
(This is explained in detail in Adam's book on Sobolev Spaces, in the section on "Duality and the spaces $W^{-m,p'}(\Omega)$"