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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 $\mathcal D'(u)$.

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For what its worth, I faintly remember from a class about topological vector spaces that the spaces of distributions are nuclear spaces, therefore they cannot be normed spaces (all normed nuclear spaces are finite dimensional). – Tim van Beek Mar 19 '11 at 9:48
I don't think that is technically the weak-$*$-topology, since $\mathcal{D}(U)$ isn't necessarily the dual space of $\mathcal{D}'(U)$. I think that only holds on bounded domains. – Brian Mar 19 '11 at 11:59
This is the most common topology on the space of distributions. Schwartz doesn't mention any other one. – Jonas Teuwen Mar 19 '11 at 13:52
Why do you want to have a stronger topology on $\mathcal D(U)$? – Dirk Mar 19 '11 at 14:49
up vote 3 down vote accepted

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.

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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.

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Can you give reference for the second paragraph? – timur Dec 6 '11 at 7:05
See my answer I gave in this question… – Vobo Dec 29 '11 at 16:04
Thanks a lot. I also found Treves' book on TVS explains everything and is very readable. – timur Jan 1 '12 at 18:59

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