# Difference between the dual space of $H^1(\Omega)$ and the dual of $H^1_0(\Omega)$

This is cross-posted on MSE: https://math.stackexchange.com/q/1596565/9464

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 $X\subset Y$.

Could anybody give a quick example in $Y\setminus X$?

• The question is somehow ill-posed, as $X$ do not naturally embed into $Y$. In general, if $F$ is a closed linear subspace of a Banach space $E$, the space $F^*$ (with the dual norm of the induced norm on $E$) is isometrically isomorphic to $E^*/ F^{\perp}$ (with the quotient norm of the norm of $E$). So any element of $E^*$ can be used as a linear functional on $F$, and this way you represent them all, only not uniquely. – Pietro Majer Jan 4 '16 at 12:01
• Thank you for your comment. What is $F^\perp$ in $E^*/F^\perp$? – Jack Jan 4 '16 at 15:31
• $T : f \mapsto \langle f,g'\rangle$ where $g \in L^2(\Omega)$ and $g'$ is its distributional derivative, then $T \in (H^1_0(\Omega))^*$ but $T \not \in (H^1(\Omega))^*$ – reuns Sep 14 '16 at 23:44

In general, $X$ does not embed into $Y$. Indeed, suppose that $\partial \Omega$ is sufficiently smooth. Then there is the trace map $\gamma_0 \colon H^1(\Omega) \to H^{1/2}(\partial\Omega)$, $u\mapsto u\bigr|_{\partial\Omega}$, which fits into a short exact sequence $$0 \longrightarrow H_0^1(\Omega) \longrightarrow H^1(\Omega) \xrightarrow{ \ \gamma_0 \ } H^{1/2}(\partial\Omega) \longrightarrow 0.$$ By duality, one gets a surjective map $X\twoheadrightarrow Y$, $\Phi\mapsto \Phi\bigr|_{H_0^1(\Omega)}$, and an embedding $H^{-1/2}(\partial\Omega)= H^{1/2}(\partial\Omega)'\to X$, $\Psi \mapsto \langle\Psi,\gamma_0 (\cdot)\rangle$. Functionals $\Psi\in H^{-1/2}(\partial\Omega)$ regarded as elements of $X$ are zero when restricted to $H_0^1(\Omega)$.
• As an elementary example, consider dimension $n=1$ and $\Omega$ the open interval $(0,1)$. Consider the functional $\Phi(u) = \int_0^1 u'(x)\,dx$ on $H^1(\Omega)$. It's not the zero functional, but its restriction to $H^1_0(\Omega)$ is zero. – Nate Eldredge Jan 4 '16 at 16:45