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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 $T$)?

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I think even for Bill Johnson one tag to ask him is sufficient, and I thus retagged to use the existing one. –  quid Jun 22 '13 at 14:12

1 Answer 1

up vote 4 down vote accepted

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 $T_j = T_i$.

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Thanks a lot for this simple funny trick! If I understand it well, it is not even necessary to take an Auerbach basis, all one needs is a basis such that the norms of $x_i$ and $x_i^*$ are bounded by one. –  Jochen Wengenroth Jun 21 '13 at 12:06
    
That's the definition of an Auerbach basis, Jochen. If you divide the first $n$ of the above $T_i$ by $n^m$ and repeat the resulting maps $n^m$ times you see that for $c$ you can take any number larger than one. –  Bill Johnson Jun 21 '13 at 13:35
    
I mean, one does not need biorthogonality. I understood that any $c>1$ does the job. Although your trick implies the series representation for the injective tensor product I admit that this representation is probably not very useful. Nevertheless I find it strange that no book mentions that. By the way, as you answered the question only 15 minutes after I posed it, I do hope that you knew this trick from somewhere else. Thanks again. –  Jochen Wengenroth Jun 21 '13 at 14:12
    
You need biorthogonality to have the $T_i$ sum to $T$, but you could take any basis and use $n^m$ operators. I learned the trick 40 years ago from Pelczynski's proof that separable spaces with the bounded approximation property are complemented in spaces with Schauder bases. His proof was much nicer than the one in my paper with Rosenthal and Zippin. –  Bill Johnson Jun 21 '13 at 14:34

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