Expressing multivariate functions as tensor product of univariate functions I am coming across many instances where multivariate functions are being represented by tensor products of univariate functions.  The basic idea seems to be as follows: Let $\phi_j$ be a univariate function in the variable $x_j$.  Then the rank one tensor $\phi_1\otimes \phi_2 \otimes \dots \otimes \phi_d$ is the multivariate function $\Phi$ defined by $$\Phi(x_1, ..., x_d) = \phi_1(x_1) \cdots \phi_d(x_d) $$ Everywhere I'm seeing this it is stated without justification, and I am wondering how one justifies it.  My attempt at justification requires fixing some bases (assuming that one can) for the $\phi_j$, i.e. suppose $\phi_j \in$ span$\{e_{\ell}\}_{\ell=1}^{p_j}$ with $\phi_j = \sum_{\ell=1}^{p_j} \alpha_{j \ell} e_{\ell}$ for some coefficients $\alpha_{j\ell}$, then by expanding $\phi_1\otimes \phi_2 \otimes \dots \otimes \phi_d$  into the basis, and applying multi-linearity, one would arrive at a sum of the sort $$\sum \alpha_{1 k_1} \cdots \alpha_{d k_d}e_{k_1} \otimes \cdots \otimes e_{k_d} $$ Is this the right idea? Perhaps there is a construction via the universal property, but I only understand this in the case of vector spaces of linear maps.  This is not a homework question, I am just interested in tensors.  Thanks!
 A: It depends on what functions you are looking at, consider a function space:
$$\mathcal F(A)=\{f:A\to\mathbb K\mid f\text{ satisfies some conditions}\}$$
If these conditions are nice enough $\mathcal F(A)$ is closed under addition and multiplication with $\mathbb K$ and you have a vector space. Examples of nice conditions are compactly supported, continuos, differentiable, bounded, integrable etc.
If you consider a function $f:A\to\mathbb K$ and $g:B\to\mathbb K$ then
$$(f\cdot g)(a,b):=f(a)g(b)$$
defines a function on $A\times B$. If your conditions are nice enough, then from $f\in\mathcal F(A), g\in\mathcal F(B)$ it follows that $f\cdot g\in\mathcal F(A\times B)$. Again, all examples I gave above are nice enough.
Well if $f_i\cdot g_i$ are in $\mathcal F(A\times B)$, so is $\sum_i f_i\cdot g_i$. Also you have:
$$(f_1+f_2)\cdot g = f_1\cdot g+f_2\cdot g\qquad f\cdot(g_1+g_2)=f\cdot g_1+f\cdot g_2\\
(\lambda f)\cdot g=\lambda(f\cdot g)=f\cdot (\lambda g)$$
So basically, the following map is well defined:
$$\iota:\mathcal F(A)\otimes\mathcal  F(B)\to\mathcal F(A\times B)$$
given on the generating set of pure tensors by $f\otimes g\mapsto f\cdot g$.
Its also simple to check, if $\iota(\sum_i f_i\otimes g_i)=\iota(\sum_i \tilde f_i\otimes \tilde g_i)$ that you already must have $\sum_i f_i\otimes g_i = \sum_i \tilde f_i\otimes \tilde g_i$, so this map is an inclusion.
This is the reason why the tensor product of the appropriate function space on $A$ with that of the function space of $B$ is the usual first approximation of that of $A\times B$.
However in most cases $\mathcal F(A)\otimes\mathcal F(B)\neq\mathcal F(A\times B)$, but it is often true (ie for $\mathcal F=C_0, L^p$) that
$$\overline{\iota(\mathcal F(A)\otimes\mathcal F(B))}=\mathcal F(A\times B)$$
ie the closure of the inclusion is the whole space, or the "first approximation" was dense. This requires a topology on the function spaces.
