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 $T$)?



series representation in injective tensor products
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 $T\_j = T\_i$. 


