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Let $u,v$ be two distribution in $\mathcal{D}'(X),\mathcal{D}'(X')$:

given tensor product $$D'(X)\times D'(X') \to D'(X\times X')\\(u,v)\mapsto u\otimes v$$

prove tensor product is a separately continuous bilinear form on $D'(X)\times D'(X') $ a similar question

The question here is what's topology on $D'(X)$ ,do we use weak topology?

I can only show sequential continuous result that is $u_n\to u$ in weak topology ,then $u_n\otimes v \to u\otimes v$ in weak topology.

How to show the continuous result does sequential continuous = continuous for $D'(X)$ with weak topology?

yi li
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  • The topology on $\mathcal D'$ is the weak topology, i.e. $u_k \to u$ if $\langle u_k, \varphi \rangle \to \langle u, \varphi \rangle$ for every $\varphi \in \mathcal D = C^\infty_c$. – md2perpe Dec 13 '20 at 14:25
  • @md2perpe Hi,here is the point,for weak topology,convergence is characterized by this,but the point is why continuous of this bilinear map is charaterized by sequential continuous for weak topology? – yi li Dec 13 '20 at 14:29
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    You can read about the topology on $\mathcal D'$ in this section of the Wikipedia article on distributions. – md2perpe Dec 13 '20 at 18:28
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    @md2perpe: before clicking on the wiki link I was expecting it to talk about the weak topology, i.e. the wrong topology, but wiki did the right thing and used the strong topology. – Abdelmalek Abdesselam Dec 13 '20 at 22:48
  • @Abdelmalek Abdesselam Hi do you have some reference book talk about theory of distribution under the assumption of Topological vector space . – yi li Dec 14 '20 at 02:57
  • There is no book I am happy with. Try this first https://math.stackexchange.com/questions/3510982/doubt-in-understanding-space-d-omega/3511753#3511753 – Abdelmalek Abdesselam Dec 15 '20 at 14:15

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