Do the operators in $B(E,F)$ separate points on the projective tensor product $F' \mathop{\tilde\otimes_\pi} E$?

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.

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.

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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