TITLE: Relationship between $C_c^\infty(\Omega,\mathbb R^d)'$ and $H_0^1(\Omega,\mathbb R^d)'$
QUESTION [5 upvotes]: 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 $H$?

REPLY [2 votes]: I think your second argument is somewhat going into the wrong direction. You want to show that $H'\subset\mathcal{D}'$, so have to show two things:
1) Given $f\in H'$, you have $f\in\mathcal{D}'$, which follows since $\mathcal{D}$ is continuously embedded into $H$ (that you still have to show).
2) If two functionals $f,g\in H'$ coincide on $\mathcal{D}$, then they are equal. This follows since $\mathcal{D}\subset H$ is a dense subspace. Note that this does not mean that every functional on $\mathcal{D}$ has an extension to $H$. It only means if there is an extension, then it has to be unique, i.e., there is at most one extension.