Question

Hi,

(Question updated)

My question is about the space of distributions of finite order $\mathcal{D}'_F$ (say on $\mathbb{R}^n$). What do we know about it ?

From in the information I gathered, it seems that the natural topology on $\mathcal{D}'_F$ is the inductive limit topology of the spaces $(\mathcal{D}'^m)$ of distributions of order $m$, or equivalently, the dual topology of $\mathcal{D}_F$ [ this space being $\mathcal{D}$ as a set, but with the coarser topology of the projective limit $(\mathcal{D}^m)$ ($C^m$ functions with compact support, this is an inductive limit of Fréchet spaces with obvious semi-norms). Note that $\mathcal{D}_F$ is strictly coarser than $\mathcal{D}$ (and strictly finer than the $\mathcal{S}$), and that $\mathcal{D}'_F$ is strictly finer than $\mathcal{D}'$ (and strictly coarser than $\mathcal{S'}$).

So, the question is: what do we now about this topology on $\mathcal{D}_F$, and its strong dual $\mathcal{D}_F'$ ? It is clearly not Frechet, but is it complete ? Montel? Barrelled ? Nuclear ? Reflexive ? More generally, do we have most of the nice properties of $\mathcal{D}'$ for $\mathcal{D}_F'$ ?

Thanks

Answer 1

Note added in edit: This was an answer to the original question which considered the topology on finite order distributions induced by the topology on all distributions.

I'm assuming that by "finite order" you mean as given on this planetmath page.

The answers are: No, Yes, Yes, No, No, No, I Think So.

In more detail:

• $\mathcal{D}_F'$ is dense in $\mathcal{D}'$ because we can approximate an arbitrary distribution by a distribution with compact support and distributions with compact support have finite order. Thus also $\mathcal{D}_F'$ is not closed in $\mathcal{D}'$.

• The obvious topology is to fix a suitable topology on each of the spaces of distributions of fixed finite order and then take the induced topology on the union. I doubt that this is the same as the induced topology, but can't say for sure. It will be the finest sensible topology on this space. (I say "it" but there could be several depending on how you topologise the spaces of fixed order, essentially you need to throw in the order into the semi-norms somehow and there may be several ways to do this.)

• $\mathcal{D}_F'$ is not complete since it is not closed in $\mathcal{D}'$. It is therefore not Montel nor reflexive.

• I'm less confident about Mackey, it having been a while since I looked at these spaces. I suspect "Yes" because it's completion is Mackey and I can't see how not being complete can complicate matters. Perhaps someone more steeped in the lore of LCTVS can help out here.

I'm curious as to the motivation for this question. Could you elaborate?

Answer 2

Thanks for your detailed answer. I am learning about topological vector spaces and distributions and in the books that I study, almost nothing is said about this space, although it seems, to me, very interesting in the following sense:

• It is big enough to contain all tempered distribution, and enough for day-to-day practice

• It has a natural grading (the notion of order) (like symbol spaces)

• It is stable under derivations

From in the information I gathered, it seems that the natural topology on $\mathcal{D}'_F$ is the inductive limit topology of the spaces $(\mathcal{D}'^m)$ of distributions of order $m$, or equivalently, the dual topology of $\mathcal{D}_F$ [ this space being $\mathcal{D}$ as a set, but with the coarser topology of the projective limit $(\mathcal{D}^m)$ ($C^m$ functions with compact support, this an inductive limit of Fréchet spaces with obvious semi-norms) ].

I guess that my question is now: what do we now about this topology on $\mathcal{D}_F$ ? It is clearly not Frechet, but is it complete ? Montel? Barrelled ? Nuclear ? Reflexive ? Same questions of course for its (strong or weak) dual.

Thanks again.