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I've seen claims that it is known that for a pair of bounded injective linear operators $T\colon X\to Y, S\colon W\to V$, their tensor product $T\otimes S\colon X \otimes_\pi W\to Y \otimes_\pi V$ need not be injective. Here $\otimes_\pi$ stands for the projective tensor product of Banach spaces.

  1. Can this happen when $T = {\rm id}_X$, the identity operator on some Banach space $X$?
  2. If so, can it happen for $T = {\rm id}_{L_1}$, the identity operator on $L_1$?

Question 2 has negative answer when $S$ is an isomorphism onto its range. Not surprisingly, the answer would be always positive for the injective tensor product.

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    $\begingroup$ For Q2, I guess we get $\operatorname{id}\otimes S : L_1 \otimes_\pi W \rightarrow L_1 \otimes_\pi V$ but then we know that $L_1 \otimes_\pi W \cong L_1(W)$ and for $V$ and so wouldn't "working with functions" show that Q2 has a negative answer (i.e. it is always injective)? $\endgroup$ May 5 at 10:16
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    $\begingroup$ For Q1: Isn't $T\otimes S=(id_Y\otimes S)\circ (T\otimes id_W)$? If the composition isn't injective then so is one factor. $\endgroup$ May 5 at 10:29

1 Answer 1

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$\require{AMScd}\newcommand{\id}{\operatorname{id}}$I use a common characterisation of the approximation property as found in e.g. Ryan's book Zbl 1090.46001.

A Banach space $X$ has the approximation property if and only if for each Banach space $Y$ (it is enough to take $Y=X^*$) the natural map $$ X \widehat\otimes Y \rightarrow X \check\otimes Y $$ is injective.

Here I write $\widehat\otimes$ and $\check\otimes$ for the completed projective, respectively, injective tensor products.

We can now answer (2) in the negative. Let $X$ have the approximation property, and let $S:W\rightarrow V$ be injective. Consider the commutative diagram $$ \begin{CD} X\widehat\otimes W @>>> X \check\otimes W \\ @V{\id\otimes S}VV @VV{\id\otimes S}V \\ X\widehat\otimes V @>>> X \check\otimes V \end{CD} $$ The map $\id\otimes S: X \check\otimes W \rightarrow X \check\otimes V$ is injective, and the horizontal arrows are injective as $X$ has AP, so $\id\otimes S: X \widehat\otimes W \rightarrow X \widehat\otimes V$ is injective. In particular $X=L_1$ has the AP, showing the negation of (2).

As Jochen Wengenroth noted, Q1 can be reduced to the $T\otimes S$ case which the OP stated has a positive answer. However, here is a concrete example, following Chapter 5, Corollary 4 of Defant and Floret Zbl 0774.46018. Let $X$ be any Banach space, and let $B_{X^*}$ be the unit ball of the dual space $X^*$, consider $\ell_\infty(B_{X^*})$ and define $j:X\rightarrow \ell_\infty(B_{X^*})$ by evaluation: $j(x) = ( \phi(x) )_{\phi\in B_{X^*}}$. Then $j$ is an isometry onto its range. We know that $\ell_\infty(B_{X^*})$ has AP so $$ X^* \widehat\otimes \ell_\infty(B_{X^*}) \rightarrow X^* \check\otimes \ell_\infty(B_{X^*}) $$ is injective. Consider now the commutative diagram $$ \begin{CD} X^* \widehat\otimes X @>>> X^* \check\otimes X \\ @V{\id\otimes j}VV @VV{\id\otimes j}V \\ X^* \widehat\otimes \ell_\infty(B_{X^*}) @>>> X^* \check\otimes \ell_\infty(B_{X^*}) \\ \end{CD} $$ The bottom arrow is injective, and the right-hand down arrow is. If $X$ does not have AP then the top arrow is not injective, and so the left-hand down arrow must fail to be injective, which gives an example of (1). (There is nothing special about $\ell_\infty$ here: any Banach space $F$ with the AP and any injection $j:X\rightarrow F$ would work.)

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