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Apr 12, 2019 at 18:49 comment added Sergei Akbarov @MeisamSoleimaniMalekan why if a functional $\varphi$ is continuous on compact sets of $L(X,Y)$ then it is continuous on $L(X,Y)$ (where $L(X,Y)$ is endowed with the compact-open topology)?
Apr 12, 2019 at 18:05 comment added MSMalekan Proposition 1.e.3 in "J. Lindenstrauss and L. Tzafriri, Classical Banach Spaces, I, Springer Verlag, Classics in Mathematics, 1996." I think is what you want.
Apr 12, 2019 at 17:05 comment added Sergei Akbarov @BillJohnson could you, please, give a more precise reference? Which statement in Lindenstrauss-Tzafriri do you mean?
Apr 12, 2019 at 16:24 comment added Bill Johnson Your conjecture is correct. This is covered in volume one of Lindenstrauss-Tzafriri. Tricky things happen when X fails the approximation property (one must distinguish nuclear operators from nuclear tensors), but that is also treated in Lindenstrauss-Tzafriri. X has the AP if and only if trace is well defined for nuclear operators. That is, if X fails the AP, then there is a nuclear representation of the zero operator for which the trace of the representation can be anything.
Apr 12, 2019 at 14:37 history edited Sergei Akbarov CC BY-SA 4.0
added 49 characters in body
Apr 12, 2019 at 14:29 history asked Sergei Akbarov CC BY-SA 4.0