Let $X$ and $Y$ be smooth manifolds. The map $\mathcal{D}'(X)\times\mathcal{D}'(Y)\to\mathcal{D}'(X\times Y)$ given by $(S,T)\mapsto S\boxtimes T$ is continuous with respect to the strong topology. Is it continuous with respect to the weak-\* topology?



Is the tensor product of distributions a continuous bilinear map with respect to the weak topology?
I will use the convention $\mathbb{N}=\{1,2,\ldots\}$ and denote by $s(\mathbb{N})$ the space of (real) sequences $(\mu\_i)\_{i\ge 1}$ of rapid decay, i.e., such that for all integer $k\ge 0$,
$$
\|\mu\|\_k:=\sup\_{i}i^k|\mu\_i|\ <\infty\ .
$$
We give it the locally convex topology defined by the norms $\|\cdot\|\_k$, indexed by integers $k\ge 0$.


The dual can be identified with the space $s'(\mathbb{N})$ of sequences $(a\_i)\_{i\ge 1}$
which grow at most polynomially in $i$.
We will write
$$
\langle \mu,a\rangle:=\sum\_i \mu\_i a\_i 
$$
for the duality pairing.


We can similarly define the two index generalizations $s(\mathbb{N}^2)$ and $s'(\mathbb{N}^2)$.
The analogue of the tensor product of distributions is the map $s'(\mathbb{N})\times s'(\mathbb{N})\rightarrow s'(\mathbb{N}^2)$ given by $(a,b)\mapsto a\otimes b$ where
$$
(a\otimes b)\_{i,j}:=a\_i b\_j\ .
$$
Saying that this bilinear map is continuous, where all spaces carry the weak-$\ast$ topology amounts to saying that $\forall \rho\in s(\mathbb{N}^2)$, $\exists p, q$, $\exists \mu^{(1)},\ldots,\mu^{(p)},\nu^{(1)},\ldots,\nu^{(q)}\in s(\mathbb{N})$, such that $\forall a,b\in s'(\mathbb{N})$,
$$
\left|\sum\_{i,j}\rho\_{i,j}a\_i b\_j\right|\le\left(\sum\_{m=1}^{p}|\langle\mu^{(m)},a\rangle|\right)
\times
\left(\sum\_{n=1}^{q}|\langle\nu^{(n)},b\rangle|\right)\ .
$$
**Lemma:** Let $\mu^{(1)},\ldots,\mu^{(p)}$ and $\mu$ be elements of $s(\mathbb{N})$.
Suppose that for all $a\in s'(\mathbb{N})$, if $\langle\mu^{(1)},a\rangle=\cdots=\langle\mu^{(p)},a\rangle=0$ then $\langle\mu,a\rangle=0$.
This hypothesis implies that $\mu$ is in the linear span of $\mu^{(1)},\ldots,\mu^{(p)}$.


For the proof, package the $\mu$'s into a $(p+1)\times\infty$ matrix and do row operations to reduce to row echelon form.


The above inequality and the lemma immediately show that the rows of the matrix $\rho\_{i,j}$ must span a finite-dimensional space. So, pick say $\rho\_{i,j}=e^{-i}\delta\_{i,j}$ and this proves that **the tensor product is not continuous for the weak-$\ast$ topology**.


Now, as I said in my comment, **the correct topology to use is the strong topology**. If all spaces carry it then saying that $(a,b)\mapsto a\otimes b$ is continuous amounts to saying


$\forall \rho\in s(\mathbb{N}^2)$, $\exists p, q$, $\exists \mu^{(1)},\ldots,\mu^{(p)},\nu^{(1)},\ldots,\nu^{(q)}\in s(\mathbb{N})$, such that $\forall a,b\in s'(\mathbb{N})$,
$$
\sum\_{i,j}\left|\rho\_{i,j}a\_i b\_j\right|\le\left(\sum\_{m=1}^{p}\sum\_i|\mu^{(m)}\_i a\_i|\right)
\times
\left(\sum\_{n=1}^{q}\sum\_j|\nu^{(n)}\_j b\_j|\right)\ .
$$


See the difference? One has traded absolute values of sums for sums of absolute values.


The statement about the strong topology is true, with $p=q=1$, because given $\rho\in s(\mathbb{N}^2)$, $\exists \mu,\nu\in s(\mathbb{N})$, such that for all $i,j$,
$$
|\rho\_{i,j}|\le |\mu\_i|\times|\nu\_j|\ .
$$


Finally, using isomorphisms with sequence spaces the above proofs also shows continuity of the tensor product operation $\mathscr{S}'(\mathbb{R})\times\mathscr{S}'(\mathbb{R})\rightarrow \mathscr{S}'(\mathbb{R}^2)$.


