Relationship between $C_c^\infty(\Omega,\mathbb R^d)'$ and $H_0^1(\Omega,\mathbb R^d)'$

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 $H$?

• Since $D \subset H$, it follows that $H'\subset D'$. Commented Jul 4, 2016 at 21:10
• @anonymus I don't think that it follows cause one set is a subset of the other. Could you explain it in more detail? Commented Jul 4, 2016 at 21:11
• Indeed, i went too fast. It is true for normed space but $D$ isn't. Its topology is quite complicated. Commented Jul 4, 2016 at 21:30

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.

• (2) is a consequence of the Hahn-Banach theorem. With "$f\in H'\Rightarrow f\in\mathcal D'$" in (1) you mean "$f\in H'\Rightarrow\left.f\right|_{\mathcal D}\in\mathcal D'$", right? Commented Jul 6, 2016 at 9:21
• @0xbadf00d: No, I don't think that (2) follows from the Hahn-Banach theorem since this does not give you uniqueness. Yes, (1) means $f\in H'\Rightarrow f|_D\in D'$. If you look at $f$ as a mapping from $H$ to $\mathbb{C}$, (1) reduces to the question of whether the embedding $D\to H$ is continuous, since then you can look at the composition $D\to H \to\mathbb{C}$, where the mapping $D\to H$ is the embedding and $H\to\mathbb{C}$ is $f$. Commented Jul 6, 2016 at 10:30

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 $L^2$.

• I only equip $\mathcal D$ with $\left\|\;\cdot\;\right\|_{\mathcal D}$ in the first part. That should be no problem. Commented Jul 4, 2016 at 21:18
• So how do you prove that $f$ is bounded in norm induced by $\langle \cdot, \cdot \rangle_H$ to apply the theorem? For example consider the second derivative of $\delta_0$, then try to prove that is bounded in the $H$ norm. (here is the mistake) Commented Jul 4, 2016 at 21:27
• You should prove that for every $\varphi \in \mathcal{D}$ you have $\varphi''(0) \leq C\|\varphi\|_H$ with $C$ independent of $\varphi$. Commented Jul 4, 2016 at 21:32
• Why should I prove that? You're right that I made a mistake in my reasoning. Is there anything else we can prove for the first part? Do you agree to the second part of my question? Commented Jul 4, 2016 at 21:37
• But if $f \in \mathcal{D}'$ then $f \notin H'$ in general. All the other people that answer are saying the same thing. Commented Jul 6, 2016 at 7:28