Let
- $d\in\mathbb N$
- $\Omega\subseteq\mathbb R^d$ be open
- $\langle\;\cdot\;,\;\cdot\;\rangle$ denote the inner product on $L^2(\Omega,\mathbb R^d)$
- $\mathcal D:=C_c^\infty(\Omega,\mathbb R^d)$ and $$H:=\overline{\mathcal D}^{\langle\;\cdot\;,\;\cdot\;\rangle_H}\color{blue}{=H_0^1(\Omega,\mathbb R^d)}$$ with $$\langle\phi,\psi\rangle_H:=\langle\phi,\psi\rangle+\sum_{i=1}^d\langle\nabla\phi_i,\nabla\psi_i\rangle\;\;\;\text{for }\phi,\psi\in\mathcal D$$
How are the topological dual spaces $\mathcal D'$ and $H'$ of $\mathcal D$ and $H$ related?
Let me share my thoughts and please correct me, if I'm wrong somewhere (and feel free to leave a comment, if everything is correct):
Let $f\in\mathcal D'$. If we equip $\mathcal D$ with the restriction $\left\|\;\cdot\;\right\|_{\mathcal D}$ of the norm induced by $\langle\;\cdot\;,\;\cdot\;\rangle_H$, then $f$ is a bounded, linear operator from $(\mathcal D,\left\|\;\cdot\;\right\|_{\mathcal D})$ to $\mathbb R$. Thus, since $\mathcal D$ is a dense subspace of $(H,\langle\;\cdot\;,\;\cdot\;\rangle_H)$, we can apply the bounded linear transform theorem and obtain the existence of a unique bounded, linear operator $F:(H,\langle\;\cdot\;,\;\cdot\;\rangle_H)\to\mathbb R$ (i.e. $F\in H'$) with $$\left.F\right|_{\mathcal D}=f\tag 1$$ and $$\left\|F\right\|_{H'}=\left\|f\right\|_{\mathcal D'}\tag 2$$ where $\left\|\;\cdot\;\right\|_{\mathcal D'}$ denotes the operator norm on $(\mathcal D,\left\|\;\cdot\;\right\|_{\mathcal D})'$.
On the other hand, if $F\in H'$ and $$f:=\left.F\right|_{\mathcal D}\;,$$ then we can show that $f\in\mathcal D'$ where $\mathcal D'$ is equipped with the usual topology. It's clear that $(2)$ is verified too.
Is there any mistake in my argumentation? And what's meant if $\nabla\pi$ with $\pi\in C_c^\infty(\Omega)'$ is claimed to be an element of $H$?
You are not allowed to do equip $ \mathcal{D}$ with the restriction $ \|\cdot\|_{\mathcal{D}}$ of the norm induced by $⟨⋅,⋅⟩_H$, because if you change the norm on $ \mathcal{D}$ you change $ \mathcal{D}' $.
For example condider $ C^{\infty}_{c}$ with the $ L^2 $ norm then the dual is $ L^2$.