# $H^{-1}$'s norm: Is this true?

Let $\Omega$ to be an open set of $\mathbb R^d$ and $P\in \mathcal D(\Omega)$.

Is $$\left \|\nabla P \right \|_{(H^{-1}(\Omega))^d}=\sum_{i=1}^{d} \sup_{\eta \in \mathcal D (\Omega)}\left( \frac{\left \langle\frac{ \partial P}{\partial x_i}, \eta \right \rangle _{H^{-1}(\Omega),H^1(\Omega)}}{\left \| \eta \right \|_{H^1(\Omega)}} \right)\;\;?$$

Thanks.

• This looks rather like a definition of $\|\nabla P\|_{H^{-1}}$. – gerw Jun 8 '17 at 11:49
• Thank's @gerw, Yeah it was an error – Motaka Jun 8 '17 at 11:54
• Where does $P$ lie? $L^2$? – GaC Jun 8 '17 at 12:33

Let's focus for simplicity on the case of a single derivative, the gradient case would follow via graph norm. If we take $P \in D'(\Omega)$ then $D_iP$ (the distributional derivative of $P$) is also in $\mathcal{D}'(\Omega)$; there is one initial important issue: does $D_iP$ have a continuous extension to $H^{-1}$, i.e can we evaluate its $H^{-1}$ norm ? The answer is generally no (think about $\delta_{0}$).
If instead $P \in \mathcal{D}(\Omega)$, then we may consider the induced distribution $T_{D_i P} : u \mapsto \int_{\Omega} u D_i P \, dx$ as a bounded linear functional on $H^{-1}\underbrace{=}_{\text{def}}(H^{1}_0)'$, where in $H^{1}_0$ we choose $\| \nabla(\cdot) \|_{0}$ as equivalent norm. The dual norm of $T_{D_i P}$ is then:
$$\| D_i P \|_{H^{-1}}= \| T_{D_i P} \|_{H^{-1}}= \sup_{v \in H^{1}_0(\Omega), v \neq 0} \left( \frac{\langle T_{D_i P}, v \rangle_{H^{-1} \times H^{1}_0}}{\| v \|_{H^{1}_0}} \right) =\sup_{\phi \in \mathcal{D}(\Omega), \phi \neq 0} \left( \frac{\langle T_{D_i P}, \phi \rangle_{H^{-1} \times H^{1}_0}}{\| \phi \|_{H^{1}_0}} \right)$$
where the last equality follows from the fact that, by defintion, $\mathcal{D}(\Omega)$ is dense in $H^{1}_0(\Omega)$ w.r.t to the topology of $H^1$; while the first equality is to remember that we identify $D_i P$ with its induced distribution.