Expressing multivariate functions as tensor product of univariate functions

Question

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!

Answer 1

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.