• $d\in\mathbb N$
  • $\Omega\subseteq\mathbb R^d$ 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 $H$?

  • $\begingroup$ Since $D \subset H$, it follows that $H'\subset D'$. $\endgroup$ – anonymus Jul 4 '16 at 21:10
  • $\begingroup$ @anonymus I don't think that it follows cause one set is a subset of the other. Could you explain it in more detail? $\endgroup$ – 0xbadf00d Jul 4 '16 at 21:11
  • $\begingroup$ Indeed, i went too fast. It is true for normed space but $D$ isn't. Its topology is quite complicated. $\endgroup$ – anonymus Jul 4 '16 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.

  • $\begingroup$ (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? $\endgroup$ – 0xbadf00d Jul 6 '16 at 9:21
  • $\begingroup$ @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 $f$. $\endgroup$ – Christian Jul 6 '16 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$.

  • $\begingroup$ I only equip $\mathcal D$ with $\left\|\;\cdot\;\right\|_{\mathcal D}$ in the first part. That should be no problem. $\endgroup$ – 0xbadf00d Jul 4 '16 at 21:18
  • $\begingroup$ 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) $\endgroup$ – user318989 Jul 4 '16 at 21:27
  • $\begingroup$ You should prove that for every $ \varphi \in \mathcal{D} $ you have $ \varphi''(0) \leq C\|\varphi\|_H $ with $ C $ independent of $ \varphi $. $\endgroup$ – user318989 Jul 4 '16 at 21:32
  • $\begingroup$ 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? $\endgroup$ – 0xbadf00d Jul 4 '16 at 21:37
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    $\begingroup$ But if $ f \in \mathcal{D}' $ then $ f \notin H'$ in general. All the other people that answer are saying the same thing. $\endgroup$ – user318989 Jul 6 '16 at 7:28

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