1
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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 $0$.

$\endgroup$
  • $\begingroup$ Thanks for pointing this out and the example. I must have miss remembered or falsely interpreted the statement, where I read it. $\endgroup$ – m.s Mar 19 '18 at 15:27

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.