2
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ So given a polynomial $\Phi(x,y) = \phi_1(x)\phi_2(y) \in L^2(X,Y)$, identified with $\phi_1 \otimes \phi_2$ (via the bilinear and linear maps you gave inducing the tensor product), in the case where, say $X = Y = [-1, 1] \subset \mathbb{R}$, we can expand $\phi_1 \otimes \phi_2$ into a basis of orthonormal polynomials and represent $\Phi$ as a multi-dimensional array as per usual? Do you know of a reference/source for proofs and studies of the injectivity, surjectivity relations you've mentioned, esp. for the $L^p$ spaces? Thanks for your help. $\endgroup$ – daniel Nov 22 '16 at 23:00
  • $\begingroup$ Sorry, I had forgotten about the question. For $L^p$ you could look at a standard generating set like continuous functions and try to see why the completion of such products will give you any function. I don't know any references, and didnt find any by quick searching. $\endgroup$ – s.harp Nov 29 '16 at 12:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.