TITLE: series representation in injective tensor products
QUESTION [6 upvotes]: All books on tensor products of Banach spaces contain the well-known theorem of Grothendieck that every element of the completed projective tensor product 
$X \tilde{\otimes}_ \pi Y$ has a representation as a series $\sum\limits_{n=1}^\infty x_n \otimes y_n$ which converges with respect to the $\pi$-norm (in an appropriate sense, this is even uniform for compact sets).
Knowing this, it is most natural to ask whether the same is true for the injective tensor product $X \tilde{\otimes}_\varepsilon Y$. 
The only thing I have found in this direction is that if $X$ and $Y$ have Schauder bases $(e_n)_{n\in\mathbb N}$ and $(f_n)_{n\in\mathbb N}$ then one can order $e_n \otimes f_m$ in a suitable way to obtain a Schauder basis of the injective (as well as the projective) tensor product. This of course answers the question and I believe that it would be enough that one of the spaces has a Schauder basis.
Moreover, it seems that the question rather easily reduces to the following problem about finite rank operators between the Banach spaces $Y^*$ and $X$: Can every finite rank operator $T$ be written as a sum $\sum\limits_{k=1}^n T_k$ of one-dimensional operators such that for all $m\le n$
$$ \| \sum\limits_{k=1}^m T_k\| \le c \| T\| $$
(where the constant $c$ is independent of $T$)?

REPLY [5 votes]: Given $T: Y\to X$ of finite rank,  let $(x_i,x_i^*)$ be an Auerbach basis (meaning they are biorthogonal and both the vectors and the biorthogonal functionals all have norm one) for the range of $T$.  For $1\le i \le n$ let $T_i y  = n^{-1} x_i^*(Ty) x_i$. For $j = kn +i$,  $1 \le k \le n$ and $1\le i \le n$, let $T_j = T_i$.