How exactly can I prove that ${\cal D}(\Omega_0)\otimes{\cal D}(\Omega_1)$ equals the linear space $M$ spanned by 'simple fucntions' $\phi\in{\cal D}(\Omega_0\times\Omega_1)$ of the form $\phi(x,y)=\phi_0(x)\phi_1(y)$ for $\phi_i\in{\cal D}(\Omega_i)$?
I mean, it makes intuitively sense to me, but it still seems hard to carry out the proof in detail. What puzzles me, given bilinear $f:{\cal D}(\Omega_0)\times{\cal D}(\Omega_1)\rightarrow U$, when I try to extend $f$ to a linear function on $M$, how can I make sure that the value of $\chi\in M$ under $f$ does not depend on the decomposition of $\chi$ into 'simple functions'?
Maybe there is some book / online resource where I can look for this?
Thanks!
A good reference on this is Friendlander and Joshi, Introduction to the Theory of Distributions. The section on tensor product is at Chapter 4, page 44. The theorem you cited seems to be largely similar to Theorem 4.3.2 in the book, and it is largely a standard $\epsilon-\delta$ type definition style proof. Correct me if I am mistaken as I am not familiar with your notation.