# Density of test functions in the space of distributions -- a clarification

Let $$U \subseteq \mathbb{R}^n$$ be open and denote by $$\mathcal{D}(U)$$ the space of all compactly supported smooth functions $$U \to \mathbb{R}$$. Let $$\mathcal{D}^\prime(U)$$ be the space of all distributions $$\mathcal{D}(U) \to \mathbb{R}$$ with the standard topology.

Given a distribution $$T$$, I would like to prove that there exists a sequence $$(\psi_n)$$ in $$\mathcal{D}(U)$$ such that $$$$\label{eq:1}\tag{\ast} \lim_{n \to \infty} \left\langle \psi_n, \varphi \right\rangle = \left\langle T , \varphi\right\rangle$$$$ for all $$\varphi \in \mathcal{D}(U)$$. I became interested in this question while the following paragraph from this Wikipedia article:

The test functions are themselves locally integrable, and so define distributions. As such they are dense in $$\mathcal{D}^\prime(U)$$ with respect to the topology on $$\mathcal{D}^\prime(U)$$ in the sense that for any distribution $$T \in \mathcal{D}^\prime(U)$$, there is a sequence $$\psi_n \in \mathcal{D}(U)$$ such that $$\left\langle \psi_n, \varphi \right\rangle \to \left\langle T, \varphi \right\rangle$$ for all $$\varphi \in \mathcal{D}(U)$$. This fact follows from the Hahn-Banach theorem, since the dual of $$\mathcal{D}^\prime(U)$$ with its weak*-topology is the space $$\mathcal{D}(U)$$.

My question is as follows: how does this follow from the Hahn-Banach theorem? I understand why $$(\mathcal{D}^\prime(U))^\ast \cong \mathcal{D}(U)$$ when the former is given the weak*-topology, but I fail to see how \eqref{eq:1} follows from the Hahn-Banach theorem.

• The "best" $C^\infty_c$ approximations to $T$ are $T_k = (k^n\phi(k.) \ast T) \phi(./k)$ where $\phi \in C^\infty_c,\phi \ge 0, \phi(0)=1,\int \phi = 1$. The approximation is uniformly continuous when restricted to distributions of order $\le M$ supported on $\|x\| \le r$. Commented Mar 19, 2019 at 10:25
• This simply shows you can't believe everything on wikipedia. The appeal to the Hahn-Banach theorem for this density result is math garbage. I agree with reun that the explicit and constructive approach with convolution is "the best" way to prove this density. Moreover, you get sequential density which is stronger than just density as pointed out by Jochen. Commented Mar 19, 2019 at 14:58
• @AbdelmalekAbdesselam Well said. Commented Feb 21 at 11:29

Your scepticism ist justified. A consequence of the Hahn-Banach theorem is that a continuous linear map $$f:X\to Y$$ between two locally convex Hausdorff spaces has dense range if and only if the transposed $$f^t: Y'\to X'$$ is injective. Applying this to the inclusion $$f:\mathscr D \hookrightarrow \mathscr D'$$ one gets that $$\mathscr D$$ is weak$$^*$$-dense in $$\mathscr D'$$ -- but in general this does not imply sequential density.
• An example with density but not sequential density is to take $\mathcal{D}$ to be the finitely supported functions $\mathbb{R} \rightarrow \mathbb{R}$ so its dual $\mathcal{D}'$ is $\mathbb{R}^{\mathbb{R}}$. By the Wikipedia argument $\mathcal{D}$ is weak-* and strong-dual dense in $\mathcal{D}'$. If $\mathcal{D}$ were sequentially dense in $\mathcal{D}'$, there would be a surjection $\mathcal{D}^\mathbb{N} \rightarrow \mathcal{D}'$. But $|\mathcal{D}| = 2^{\aleph_0}$, so $|\mathcal{D}^{\mathbb{N}}| = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0} < 2^{2^{\aleph_0}} = |\mathbb{R}^\mathbb{R}|$ Commented Feb 21 at 12:20
• Can't edit, but I slightly oversimplified my last comment when making it fit the required number of characters - it should of course be a surjection from a subset of $\mathcal{D}^{\mathbb{N}}$ onto $\mathcal{D}'$. Commented Feb 22 at 6:41