I aksed this question on Math Stack Exchange 6 days ago, with no answer: https://math.stackexchange.com/q/4875445/1297919

Let $X$ and $Y$ two Banach spaces and let $X\otimes Y$ their tensor product. Let $A(u)$ be the collection of all finite sets of simple tensors $\{x_1\otimes y_1,\dots ,x_n\otimes y_n\}$ such that:

  1. $u=\sum_{i=1}^n x_i\otimes y_i,$

  2. There is no subset with at least two elements of $\{x_1\otimes y_1,\dots ,x_n\otimes y_n\}$ such that the sum of its elements is a simple tensor.

Is it true that for every $u\in X\otimes Y$ we have $\max\{\text{card}\, s:s\in A(u)\}<\infty$?


1 Answer 1


Let $a_0,a_1,a_2\dots$ be linearly independent elements of $X$ and $b_0,b_1,b_2\dots$ be linearly independent elements of $Y$. Let $u=a_1\otimes b_1-a_0\otimes b_0$, and let $$v_i=a_i\otimes (b_i-b_{i+1}),\quad w_i=(a_{i}-a_{i+1})\otimes b_{i+1},$$ $$x_i=a_i\otimes(b_i-b_0),\quad y_i=(a_i-a_0)\otimes b_0.$$ Then $$v_1+w_1+v_2+w_2+\dots+v_{n-1}+w_{n-1}+x_n+y_n= u$$ but no subset of $\{v_1,w_1,v_2,w_2,\dots,v_{n-1},w_{n-1},x_n,y_n\}$ of size at least two adds up to a simple tensor. So the answer to your question is no.

  • $\begingroup$ Thanks! Do you think that it's possibile to have an example with this property where the sets $S_n$ of simple tensors that you use at stage $n$ have empty intersection, that is $\cap_n S_n=\emptyset$? $\endgroup$ Commented Mar 13 at 11:01
  • $\begingroup$ Yes, a variation on this example should do that for you. $\endgroup$ Commented Mar 13 at 11:36

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