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I'm following the proof of the theorem 4.3.3 p. 47 of "Introduction to the distributions theory" by Friedlander and Joshi. We have the following identity

\begin{align*} \textbf{(1)} \displaystyle \langle u \otimes v, \varphi \rangle = \langle v(y), \langle u(x), \varphi(x,y) \rangle \rangle = \langle u(x), \langle v(y), \varphi(x,y) \rangle \rangle \end{align*}

where $\varphi \in \mathcal{D}(X\times Y)$. Ok I understand that this identity is worth. However, the same theorem, there is the next point

$\textbf{(4)}$ The tensor product is a separately continuous bilinear form on $\mathcal{D}'(X) \times \mathcal{D}'(Y)$

and in the proof it says that $(4)$ is immediate consequence of $(1)$. Sincerely, I did not understand what it means "separately" and how $(1)$ implies $(4)$.

Thank you for each reply.

user288972
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2 Answers2

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Separately - in this context - means "with respect to each argument, assuming that other argument is fixed".

Since $D(X\times Y)$ contains $D(X)\times D(Y)$, the same definition as in (1) applies to $D'(X)\times D'(Y)$, and therefore you can use the form in (1) to prove continuity in (4).

TZakrevskiy
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This is a cheap version of the theorem. In fact, this operation is continuous (not just separately) if one uses the strong topology on $\mathcal{D}'(X)$, $\mathcal{D}'(Y)$ and $\mathcal{D}'(X\times Y)$.

Abdelmalek Abdesselam
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