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 $T$)?

onetag to ask him is sufficient, and I thus retagged to use the existing one. $\endgroup$ – user9072 Jun 22 '13 at 14:12