# tensor product of locally convex spaces

Let $X$ and $Y$ be locally convex vector spaces, over $\mathbb{C}$, equipped with directed families of seminorms $\mathcal{P}_X$ and $\mathcal{P}_Y$ respectively inducing the topologies. For each pair $(p,q) \in \mathcal{P}_X \times \mathcal{P}_Y$ define the seminorm $p \otimes q$ on $X \otimes Y$ by $(p \otimes q)(z) = \inf{\sum_{i = 1}^{n}{p(x_i)q(y_i)}}$, the infimum being taken over all ways of writing $z = \sum_{i = 1}^n{x_i \otimes y_i}$. Equipped with the induced topology, the algebraic tensor product $X \otimes Y$ becomes a locally convex space and a tensor product in this category (in the sense that continuous linear maps out of $X \otimes Y$ are in bijection with continuous bilinear maps out of $X \times Y$, within the category locally convex spaces).

I would like to prove the following: For every tensor $z \in X \otimes Y$ of rank $n \geq 1$ one has the equality $(p \otimes q)(z) = \sum_{i = 1}^n{p(x_i) q(y_i)}$ for all $x_i \in X$, $y_i \in Y$ such that $z = \sum_{i = 1}^n{x_i \otimes y_i}$.

ADDED: As Jochen points out in his answer below, this is hopeless for $n \geq 2$... Oops.

I know how to prove this when $n =1$: Abbreviate $x = x_1$, $y = y_1$. One the one-dimensional subspaces $\mathbb{C} x \subset X$ and $\mathbb{C} y \subset Y$ consider linear functionals $\phi, \psi$ satsifying $\phi(x) = p(x)$, $\psi(y) = q(y)$. By Hahn-Banach, they can be extend to continuous linear functionals $\Phi : X \rightarrow \mathbb{C}$, $\Psi : X \rightarrow \mathbb{C}$ satisfying $|\Phi(a)| \leq p(a)$ for all $a \in X$ and $|\Psi(b)| \leq q(b)$ for all $b \in Y$. Then, for every way of writing $x \otimes y = \sum_{i = 1}^n{a_i \otimes b_i}$, we have $$p(x)q(y) = \phi(x)\psi(y) = \Phi(x) \Psi(y) = (\Phi \otimes \Psi)(x \otimes y) = (\Phi \otimes \Psi)( \sum_{i = 1}^n{a_i \otimes b_i}) \\ = \sum_{i = 1}^n{\Phi(a_i) \Psi(b_i)} = \left| \sum_{i =1}^{n}{\Phi(a_i) \Psi(b_i)} \right| \leq \sum_{i =1}^{n}{|\Phi(x_i)| |\Psi(y_i)|} \leq \sum_{i = 1}^{n}{p(x_i) q(y_i)}$$ From the definition of infimum, we deduce $p(x)q(y) \leq (p \otimes q)(x \otimes y)$, and then equality since the remaining inequality follows directly form the definitions.

I can't seem to figure out how to generalize this method to $n \geq 2$. Any help would be much appreciated.

ADDED: Definition An element $z \in X \otimes Y$ has rank $n \geq 0$ if there exit $x_1, \dots, x_n \in X$ and $y_1, \dots, y_n$ such that $z =\sum_{i =1}^{n}{x_i \otimes y_i}$ and if $n$ is minimal with this property, i.e. if there does not exist $m < n$ and elements $a_1, \dots, a_m \in X$, $b_1, \dots, b_m \in Y$ such that $z = \sum_{i =1}^{m}{a_i \otimes b_i}$.

This definition implies that whenever we express an element $z \in X \otimes Y$ of rank $n \geq 0$ as $z = \sum_{i =1}^n{x_i \otimes y_i}$, the $x_i$'s are necessarily linearly independent (similarly for the $y_i$'s).

Rank 1 tensors are VERY particular. For $n\ge 2$ the statement is just wrong. Take $x\in X$, $y\in Y$ of norm $1$ and very small but linearly independent vectors $e\in X$ and $f\in Y$ and consider the rank 2 tensor $$z=x\otimes y +(-x+e)\otimes (y+f).$$ For this representation $\sum\limits_{i=1}^2 p(x_i)q(y_i)$ is close to $2$. But $z$ has another representaion $z= e\otimes y +(-x+e)\otimes f$ for which $\sum\limits_{i=1}^2 p(x_i)q(y_i)$ is close to $0$.