TITLE: Proving continuity on spaces of distributions?
QUESTION [6 upvotes]: Let $\mathcal{D}'(\Omega)$ be the space of distributions on an open set $\Omega$, and $\mathcal{E}'(\Omega)$ the compactly supported ones.
When you have a linear operator $T:\mathcal{D}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$ or $T:\mathcal{E}'(\Omega)\rightarrow\mathcal{E}'(\Omega)$, I often find it easy to prove that convergent sequences get mapped to convergent sequences (in the weak topology), but proving continuity with respect to the weak topology is much harder. Since these spaces are not first-countable (I think), one must work with a local basis around 0.
For example, if you have a pseudodifferential operator $P:\mathcal{E}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$ which is properly supported, then it continuously extends to a (necessarily unique) operator $P:\mathcal{D}'(\Omega)\rightarrow\mathcal{D}'(\Omega)$ and maps $\mathcal{E}'(\Omega)$ continuously into $\mathcal{E}'(\Omega)$. I was able to prove the first statement, but I was able to prove the second statement only in terms of sequential continuity. Note that the weak topology of $\mathcal{E}'(\Omega)$ is not the subspace topology that it inherits from the weak topology of $\mathcal{D}'(\Omega)$ (I think), so this is not trivial.
In many textbooks involving distributions, authors don't seem to be careful about this. They give an argument proving sequential continuity and claim that it is continuous in the more general sense. Is there a general scheme for converting sequential continuity arguments to actual continuity arguments for these spaces? Or can you point me to a detailed proof of the aforementioned fact about proper PDOs?

REPLY [4 votes]: Every linear mapping $T:\mathcal D'\to \mathcal D'$ which is $\sigma(\mathcal D',\mathcal D)$-sequentially continuous is $\sigma(\mathcal D',\mathcal D)$-continuous.
Perhaps, this is the same as Peter Michor's answer, but let me try to explain that the major point is the bornologicity of $\mathcal D'$ endowed with the strong topology
$\beta (\mathcal D',\mathcal D)$ (uniform convergence on the bounded subsets of $\mathcal D$) which follows e.g. from a beautiful theorem of Laurent Schwartz (the strong dual a every complete Schwartz space is bornological).
Moreover, since $\mathcal D$ is barrelled, the strong and the weak-$*$ dual have the same bounded sets.
If now $T: (\mathcal D',\sigma(\mathcal D',\mathcal D)) \to (\mathcal D',\sigma(\mathcal D',\mathcal D))$ is sequentially continuous then every bounded sequence of $\mathcal D'$ is mapped to a bounded sequence and hence $T$ is bounded on bounded sets
of $\mathcal D'$. As $(\mathcal D',\beta(\mathcal D',\mathcal D))$ is bornological
$T:(\mathcal D',\beta(\mathcal D',\mathcal D)) \to (\mathcal D',\beta(\mathcal D',\mathcal D))$ is continuous and therefore
$T:(\mathcal D',\sigma(\mathcal D',\mathcal D''))\to (\mathcal D',\sigma(\mathcal D',\mathcal D''))$ is continuous. By reflexivity, you get the desired weak-$*$ continuity.
In general, you are perfectly right with your skepticism against careless arguments
involving sequential continuity. For example, it may very well be that for a linear partial differential operator with constant coefficients $T=P(D): \mathcal D'(\Omega) \to \mathcal D'(\Omega)$ the inverse of $P(-D): \mathcal D(\Omega) \to \mathrm{Im} (P(-D))$ is sequentially continuous without being continuous (the former condition is equivalent to $P$-convexity, the latter to strong $P$-convexity)