# Tensors as linear combinations of pure tensors.

Let $V$ be an n-dimensional real vector space, consider the space $F(V^p)$ of real functions on the p-fold cartesian product $V^p$ and its subspace $(V^{*})^p$ of multilinear functions (i.e. covariant tensors).

I know that every $f\in (V^*)^p$ can be written as a finite sum of pure tensors, i.e. (for simplicity I take $p=2$)

$f(v,w)= f_1(v)g_1(w) +\dots f_N(v) g_N(w)$

for some positive integer $N$ and $f_i,g_i\in V^*$. (In fact $N$ can be taken independent of $N$, the dimension of $(V^*)^p$ being finite).

Does a similar statement hold for $F(V^p)$ even if the latter is an infinite dimensional vector space? I.e. (in the case $p=2$ for simplicity) ), given $f\in F(V^2)$ do there exist a positive integer $N$ and functions $f_i,g_i:V\rightarrow \mathbb R$ for $i=1,\dots, N$ such that

$f(v,w)= f_1(v)g_1(w) +\dots f_N(v) g_N(w)$?

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You need to first think about the definitions: what does it mean to take the tensor product of two (possibly infinite dimensional) vector spaces? Perhaps refer to Wikipedia and also this previous question. –  Willie Wong Mar 29 '12 at 14:16
In particular, in the infinite dimensional setting the expression $(V^*)^{\otimes p} = (V^{\otimes p})^*$ is no longer true. So your assertion in the first paragraph "multilinear functions (i.e. covariant tensors)" already runs into difficulty. –  Willie Wong Mar 29 '12 at 14:18
@Willie Wong. I'm not sure I understand your comments. Maybe I was not clear and there is a misunderstanding: the vector space $V$ is always FINITE dimensional, and i never take tensor products of infintie dimensional spaces, also not in the statement I want to prove/disprove. I just oberved that what I call $F(V^p)$ is an infinite dimensional vector space. –  Hans Mar 29 '12 at 14:44
Ah I see what you meant. For clarification: are you looking for a decomposition of $F(V^p)$ as a product of $F(V)$? Then consider the following: let $f:V\times V \to \mathbb{R}$ be given by $f(v,w) = 0$ unless $v=w$, in which case $f(v,w) = 1$. This function cannot be written as a finite linear combination of functions of the form $g(v)h(w)$. In fact, you cannot even get by with countable combinations. –  Willie Wong Mar 29 '12 at 14:54
@Willie Wong. Yes, this is was I meant. And your counterexample gives an answer (which I would accept if you make it an official answer). Many thanks. –  Hans Mar 30 '12 at 13:00
Consider the following: let $f : V\times V\to \mathbb{R}$ be given by $f(v,w) = 0$ when $v\neq w$, and $f(v,w) = 1$ if $v = w$. (This is the generalisation of the Kronecker $\delta$ symbol, and can be also thought of as the characteristic function of the diagonal set in $V\times V$.) This function cannot be written as a finite (or even countably infinite) linear combination of functions of the form $g(v)h(w)$. (Since $V$ is a $\mathbb{R}$ vector space and contains uncountably many points.)
This shows that $F(V^2)$ cannot be decomposed as finite combinations of $F(V)$.