7
$\begingroup$

Let $E$ and $F$ be Banach spaces, and let $\mathfrak L_{co}(E,F)$ denote the space of bounded linear operators $E \to F$ equipped with the topology of uniform convergence on the absolutely convex compact subsets of $E$.

Defant and Floret [DF93, §5.5] point out that there is a natural linear map $$ D_F : F' \mathop{\tilde\otimes_\pi} E \to (\mathfrak L_{co}(E,F))' $$ which is surjective. (No continuity claims are made.) Furthermore, they prove [DF93, §5.7] that $D_F$ is injective if $F$ is reflexive or if $F'$ or $E$ has the approximation property, so in that case one has an isomorphism $F' \mathop{\tilde\otimes_\pi} E \cong (\mathfrak L_{co}(E,F))'$ of vector spaces. They conclude with:

We do not know whether or not $D_F$ is always injective.

Question. Has this since been settled? Can someone provide (a reference to) a proof or a counterexample?


Some background: $D_F$ is defined by taking the natural map $$ \Phi_F : \mathfrak L(E,F) \stackrel{1}{\hookrightarrow} \mathfrak L(E,F'') \stackrel{1}{\cong} (E \mathop{\tilde\otimes_\pi} F')' \stackrel{1}{\cong} (F' \mathop{\tilde\otimes_\pi} E)'. $$ It is shown that $\Phi_F$ is continuous as a map $\mathfrak L_{co}(E,F) \to [(F' \mathop{\tilde\otimes_\pi} E)',\text{weak-$*$}]$, and then $D_F$ is defined as the adjoint of this map. (I guess one could deduce some continuity property of $D_F$ from this, but the authors steer clear of that, presumably to focus on questions related to Banach spaces.)

Another way to interpret the question is this: the map $D_F$ (or $\Phi_F$) gives rise to a bilinear map $(F' \mathop{\tilde\otimes_\pi} E) \times \mathfrak L(E,F) \to \mathbb{F}$. One readily verifies that this is simply the natural map $$ \left(\sum_{n=1}^\infty y_n' \mathop{\otimes} x_n \, , \, T\right) \mapsto \sum_{n=1}^\infty y_n'(Tx_n). $$ Since we have $\mathfrak L(E,F) \stackrel{1}{\hookrightarrow} (F' \mathop{\tilde\otimes_\pi} E)'$, it is clear that $F' \mathop{\tilde\otimes_\pi} E$ separates points on $\mathfrak L(E,F)$. The question whether $D_F$ is injective is equivalent to the question whether $\mathfrak L(E,F)$ separates points on $F' \mathop{\tilde\otimes_\pi} E$ (so that the bilinear map becomes a dual pairing).

Yet another equivalent formulation: is the image of $\Phi_F$ weak-$*$ dense in $(F' \mathop{\tilde\otimes_\pi} E)'$?


References.

[DF93] A. Defant, K. Floret, Tensor Norms and Operator Ideals (1993), Mathematics Studies 176, North-Holland.

$\endgroup$
3
$\begingroup$

It follows from this answer of Bill Johnson that it can already fail if $E = F$, so the question has been settled in the negative.

$\endgroup$
  • 1
    $\begingroup$ (Disclaimer: I did search for related questions before asking, but only noticed the linked answer after meticulously crafting a list of all problems concerning the approximation property for a question on meta: Suggested tag: approximation-property. I'm pretty sure I wouldn't have found it otherwise. Oh well.) $\endgroup$ – J. van Dobben de Bruyn Aug 29 '19 at 21:03
  • 2
    $\begingroup$ Don't feel bad. I spent a couple of hours, with no success, trying to build a counterexample for your question. I had completely forgotten the linked question and my answer to it. $\endgroup$ – Bill Johnson Aug 29 '19 at 22:01
2
$\begingroup$

This question was settled in 2012 by Petr Hájek and Richard J. Smith [HS12]. They prove the following beautiful result (reformulated here to match the notation from the question).

Theorem (cf. [HS12, Theorem 2.5]). Let $F$ be a Banach space with the AP. Then the following conditions are equivalent:

  • $F'$ has the AP.
  • For every Banach space $E$, the map $F' \mathbin{\tilde\otimes_\pi} E \to (\mathfrak L_{co}(E,F))'$ is injective.
  • The map $F' \mathbin{\tilde\otimes_\pi} F'' \to (\mathfrak L_{co}(F'',F))'$ is injective.

As there are known examples of Banach spaces with the AP whose dual fails the AP, the question is settled in the negative.


References.

[HS12]: Petr Hájek, Richard J. Smith, Some duality relations in the theory of tensor products, Expositiones Mathematicae, volume 30 (2012), issue 3, pages 239–249. https://doi.org/10.1016/j.exmath.2012.08.004

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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