I often stumbled across (variations of) the notation $f\otimes g$ for $(x,y)\mapsto f(x)g(x)$.
If $f\in V^*$, $g\in W^*$ for vector spaces $V,W$ over ${\bf K}$ then $f\otimes g:x\otimes y\mapsto f(x)g(y)$ is the standard definition of the tensor product of linear maps where we associate ${\bf K}\otimes{\bf K}$ with ${\bf K}$ in the canonical way.
But sometimes I read this in contexts where $f,g$ are not linear functionals, e.g. $f,g\in{\cal C}^\infty(\Omega)$ for $\Omega\subseteq{\bf R}^n$, here I always thought $f\otimes g$ denotes an element in ${\cal C}^\infty(\Omega)\otimes{\cal C}^\infty(\Omega)$, but I could not prove that ${\cal C}^\infty(\Omega)\otimes {\cal C}^\infty(\Omega)$ is a subset of ${\cal C}^\infty(\Omega\times\Omega)$, i.e. the space where $(x,y)\mapsto f(x)g(y)$ lives in.
So is this just a notational convention derived from the case of linear functionals, or is there something more subtle going on which I'm not able to see yet?
Just before answering the question I want to restate my problem. I often see theorems in the form also described here where one shows that $F(X)\otimes F(Y)$ lies dense in $F(X\times Y)$ for some functor $F$ associating spaces with corresponding function spaces but most of the time they begin like 'define $F(X)\otimes F(Y)$ as the subspace generated by $f\otimes g:=(x,y)\mapsto f(x)g(y)$,' i.e. they leave it open if we deal with an actual tensor product here or not, and this information is also not needed at all to successfully state and prove such a theorem. Still it bothered me and I looked for the (apparently much easier theorem) that in such cases the subspace generated by the maps $(x,y)\mapsto f(x)g(y)$ is indeed a tensor product of $F(X)$ and $F(Y)$.
What helped me was a different defining property (alternative to the usual universal property) found in Francois Treves' book on p. 403, namely $(M,\phi)$ is a tensor product of $E,F$ if $\phi:E\times F\rightarrow M$ is bilinear, the image of $\phi$ spans $M$ and $E$, $F$ are $\phi$-linearly disjoint, meaning if $\sum_{i<n}\phi(x_i,y_i)=0$ and the $x_i$ are independent then $y_i=0$ for $i<n$.
Using this property it is very easy to check the claim in a lot of different cases. Treves gives also a nice example which can be applied most of the time, namely if $X,Y$ are sets and $E,F$ linear spaces of ${\bf K}$-valued functions (he considers only complex values but it should make no difference) on $X$ resp. $Y$ then the space spanned by elements $(x,y)\mapsto f(x)g(y)$ for $f\in E$, $g\in F$ gives a tensor product of $E$ and $F$.