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Let $\Omega \subseteq \mathbb{R}^n$ be a nonempty open set, and $\mathcal{D}(\Omega)$ the space of test functions (that is infinitely differentiable functions $f:\Omega \rightarrow \mathbb{C}$ with compact support contained in $\Omega$), with the usual topology defined through inductive limit of Fréchet spaces (see Distribution or any good book which deals with distributions as Rudin, Functional Analysis, or Reed and Simon, Methods of Mathematical Physics, Volume I or the wonderful Schwartz, Théorie des Distributions).

Now, let us recall that a topological space $X$ is called a Fréchet-Urysohn space if for every $A \subseteq X$, the closure of $A$ coincides with the sequential closure $[A]_{seq}$ of $A$, which is defined as \begin{equation} [A]_{seq} = \{ x \in X : \exists (a_j)_{j=0}^{\infty} : a_j \rightarrow x \textrm{ and } a_j \in A \textrm{ for } j=0,1,2,\dots \}. \end{equation} A set $A \subseteq X$ for which $A= [A]_{seq}$ is called sequentially closed. Note that every closed set $A$ is sequentially closed. If also the converse is true, that is if every sequentially closed set turns out to be closed, the space $X$ is called a sequential space (see Sequential Space for more details and references about sequential spaces). Clearly, every Fréchet-Urysohn space is a sequential space, but the converse is not true (see the post Understanding two similar definitions).

With this terminology we may ask: is $\mathcal{D}(\Omega)$ a Fréchet-Urysohn space? Is it a sequential space?

The answer to these two questions is negative, as I will show in my answer below. I posted the question here only to share this result with the community of math.stackexchange.com since I coud not find the answer in any book I consulted.

Martin Sleziak
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Maurizio Barbato
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  • If this space is not sequential, then continuous functionals on it are in general different from the sequentially continuous ones. However, $\mathcal{D}'$ is usually defined through sequential continuity. Does this mean that the actual topological dual of $\mathcal D$ is different from $\mathcal{D}'$, and if so, why don't we use it as the space of distributions? – Alex Bogatskiy Mar 31 '17 at 16:57
  • It seems that the sequential characterization only holds for linear functions. @Maurizio_Barbato I wish I had more time to go into all the details, but I still have a doubt about your answers (which I do not fully understand yet). Nevertheless, it seems contradictory to what is written on wikipedia about sequential spaces, namely that they are stable under inductive limits. Each $\mathcal{D}_K$ is a normed space so sequential but you are saying the inductive limit is not... – Noix07 Mar 17 '19 at 17:20
  • A way to accommodate the apparent contradiction: the inductive limit in the category of locally convex topological vector space cannot be the inductive limit in the category of sequential spaces. Maybe there are more sequential TVS compatible with the topologies of $\mathcal{D}_K$ or maybe fewer. Maybe unrelated but this article seems indeed to study different locally convex top with the same convergent sequences. – Noix07 Mar 19 '19 at 15:46
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    @Noix07 Dear Noix07 I have never studied category theory, so I cannot help you with your doubts. As for the result proved in this post, namely that $\mathcal{D}(\Omega)$ is not a sequential space, you can also take a look at the original proof given by Shirai in his work Sur les topologies des espaces de L. Schwartz. Anyway, I suggest to you to deepen your knowlewdge of functional analysis and distribution theory before tackling delicate issues as those dealt with in this post. Good luck with your study! – Maurizio Barbato Mar 21 '19 at 19:51
  • Thanks for the reference. It uses some old terminology that one does not even find in Bourbaki (Espaces vectoriels topologiques) but which is mentionned in the mémoires by Laurent Schwartz: "I defined exactly, in Grenoble, the "true topology" really corresponding to the "pseudo-topology" of $\mathcal{D}$, which is, what later, Dieudonné and me would call, in 1946, an inductive limit. One cannot be content with the pseudo-topology, one needs a topology to apply the Hahn-Banach theorem to study subspaces of $\mathcal{D}$". cf. pseudo-top – Noix07 Mar 22 '19 at 10:30
  • Here is an answer with a reference to another proof that $\mathcal{D}$ is not sequential. A second reference can be found in the comments of that answer – Noix07 Mar 22 '19 at 10:57

1 Answers1

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I will make use here of the notation introduced in my previous post Topology of the space $\mathcal{D}(\Omega)$ of test functions. We shall show that $\mathcal{D}(\Omega)$ is not a sequential space, so that the answer to both questions is negative, as announced.

Take $V$ as in my answer to the post Topology of the space $\mathcal{D}(\Omega)$ of test functions and let $A$ be the complement of $V$. Then the argument given in that answer shows that $0$ is a limit point of $A$, so $A$ is not closed. Anyhow, $A$ is sequentially closed, as we shall now show.

Suppose that $f \in V$ and that $(f_j)$ is a sequence in $\mathcal{D}(\Omega)$ converging to $f$. Then by the characterization of converging sequences in $\mathcal{D}(\Omega)$ (see e.g. Theorem (6.5) in Rudin, Functional Analysis, 2nd Edition), we know that:

(i) there is a compact set $K$ contained in $\Omega$, such that the support of $f_j$ is contained in $K$ for all $j=0,1,2,\dots$,

(ii) for every $\epsilon > 0$ and every nonnegative interger $N$ there exists a nonnegative integer $m$ such that $\left| \left| f_j - f \right| \right|_N < \epsilon$ for all $j \geq m$.

Now, since $V \cap \mathcal{D}_K \in \tau_k$, there exists $\epsilon > 0$ and a nonnegative integer $N$ such that the set \begin{equation} B = \{ g \in \mathcal{D}_K : \left| \left| g - f \right| \right|_N < \epsilon \} \end{equation} is contained in $V \cap \mathcal{D}_K$. Then, if $m$ is the nonnegatve integer whose existence is stated in (ii), we conclude that $f_j \in V$ for all $j \geq m$. So there is no sequence $(f_j)$ in $A$ converging to $f$.

QED

NOTE. From NOTE (2) in my answer to the post Topology of the space $\mathcal{D}(\Omega)$ of test functions, we know that $A$ is dense in $\mathcal{D}(\Omega)$, and since $A$ is sequentially closed, we can conclude that $A$ is an example of a dense subset of $\mathcal{D}(\Omega)$ which is not sequentially dense in $\mathcal{D}(\Omega)$.

Maurizio Barbato
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  • I'm not sure I understand the business with $A$. What I do understand is that although $V$ is not open (as proved in the other answer), it is sequentially open. – Noix07 Mar 19 '19 at 15:32