# Tensor product notation for $(x,y)\mapsto f(x)g(y)$

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?

• There is for example this question where this notion is also used, but the author explicitly defines $f\otimes g$, i.e. does not presuppose anything about existing tensor products as far as I can tell... – fweth Feb 19 '15 at 20:38
• Also there is this passage in Treves' Topological Vector Spaces, Distributions, Kernels on page 409 where he talks about the tensor product ${\cal C}^k(X)\otimes{\cal C}^l(Y)$ for $X,Y$ subsets of ${\bf R}^n,{\bf R}^m$ respectively and notes that these are spaces of functions defined in $X\times Y$ and I just can't bring this together with the universal property of tensor products -.- – fweth Feb 19 '15 at 20:58
• I think I could prove it now using the techniques from Treves book, will post result tomorrow. – fweth Feb 19 '15 at 22:05
• When I have seen $X \times Y$ in the context of Banach spaces, then this usually denotes the completion of the algebraic tensor product with respect to some norm. There are several different natural notions of what this norm could be: en.wikipedia.org/wiki/Banach_space#Tensor_product. – Stephen Montgomery-Smith Feb 19 '15 at 23:03
• There could indeed be something more going on: the tensor product can be defined in different categories so that it has to be an object of the category. I've heard that it's quite complicated for Banach spaces (or algebra i don't know) because there are indeed many possible norm on the "algebraic" tensor product (considered only as a vector space). Here, I would say that the tensor product of function is that of commutative associative algebras. i also wanted to clarify this kind of details and so far i stumbled on the so called "exponential law" for maps and closed monoidal or cartesian categ – Noix07 Mar 22 '15 at 12:52

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 $F$.