$\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m) \simeq \mathcal S(\mathbb R^{n+m})$?

Question

If $S(\mathbb R^n)$ is the Scwartz space of smooth rapidly decaying functions equipped with the topology generated by the family of semi-norms $$\mathcal N_p (\varphi)= \sum_{|\alpha|, |\beta| \leq p} \sup_{x\in \mathbb R^n} |x^\alpha \partial^\beta \varphi (x) |\, ,$$ Is it true that $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m)\simeq \mathcal S(\mathbb R^{n+m})$, where $\mathcal S(\mathbb R^n) \hat \otimes_\pi \mathcal S(\mathbb R^m)$ is the completion of $\mathcal S(\mathbb R^n) \otimes \mathcal S(\mathbb R^m)$ for the topology $\pi$ defined by the family of semi-norms $$\mathcal N_{p,q} (\varphi\otimes \psi)=\mathcal N_p(\varphi)\mathcal N_q (\psi)\, .$$

Answer 1

Yes, if you understand the tensor product topology in the right way. Since the spaces are nuclear, inductive and projective tensor products coincide. The result is theorem 51.6 of Treves: Topological vector spaces, distributions, and kernels.

Added later:

Attention: The description of seminorms on the tensor-product given in the question is not sufficient to specify a locally convex topology on the tensor product. There are many satisfying this description; all between the the projective one and the the inductive one and even more. See the source I have given, or many other books.