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.

  • 2
    $\begingroup$ 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, 2022 at 10:16
  • 5
    $\begingroup$ 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, 2022 at 10:29

1 Answer 1


$\require{AMScd}\newcommand{\id}{\operatorname{id}}$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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.