Topologies on the space $\mathcal D'(U)$ of distributions

Question

In my analysis lecture I am given a topology on the space of distributions as follows:

Let $u_k$ be a sequence in $\mathcal D'(u)$, $u \in \mathcal D'(u)$. We say $u_k \rightarrow u$, if $\forall \phi \in \mathcal D(u) : u_k(\phi) \rightarrow u(\phi)$.

This is the weak-$*$-topology on $\mathcal D'(u)$. It seems lecturers don't care too much about the topology of $\mathcal D'(u)$, hence I wonder whether there are stronger topologies on $\mathcal D'(u)$.

Answer 1

Certainly there exist stronger topologies on distributions, but as a practical matter the weak-* definition is the one that is interesting and I assume that was the direction of your question. There isn't the usual norm topology available on $\mathcal D(U)$, and per Tim's comment do not have a different norm topology either.

$\mathcal D(U)$ is a pretty strict space to be in and to converge in, so it isn't very demanding to be a distribution. The hard work is all put on the test functions, so to speak. Although there is a certain amount of interesting things you can do with distributions, practically distributions are a stepping stone for getting to more interesting spaces, such as using their differentiability properties to define Sobolev spaces.

Answer 2

Some remarks:

$D(U)$ is reflexive, even a Montel space, so the dual of $D'(U)$ with weak or strong topology is again $D(U)$ [This is in contrast to Brian's remark].

A linear functional on $D(U)$ is continuous (i.e. a distribution) if and only if it is sequentially continuous. This is remarkable, as the space of test functions $D(U)$ is not metrizable, so sequential continuity is usually not sufficient.

A sequence of distributions is weakly convergent if and only if it is strongly convergent [i.e. uniformly on bounded subsets of $D(U)$].

The last remark is why usually only the weak topology is known. And Schwartz proved and mentioned this consequence quite often in his book.