Let $E,F$ be Banach spaces and let $A\subset K(E,F)$ be a subset of the space of compact operators from $E$ to $F$.
A result by Kalton states that $A$ is weakly compact if and only if $A$ is WOT\* compact (here WOT\* denotes the dual weak operator topology, i.e. the topology defined by the functionals $K(E,F)\ni T\mapsto e''(T^\*f')$, for $e''\in E^{\*\*}$, $f'\in F^\*$):


N. Kalton, "Spaces of compact operators", Math. Ann. 208, 267--278 (1974) .


On the other hand, the Eberlein-Smulian theorem tells us that weak compactness is equivalent to weak sequential compactness.


Question: Is WOT\* compactness in $K(E,F)$ equivalent to WOT\* sequential compactness? (maybe under some extra assumptions)



Weak sequential compactness on the space of compact operators
The key here is the isometric embedding of $K(E,F)$ into the space of continuous functions on the compact space $M=B\_{E^{\*\*}}\times B\_{F^\*}$.


Suppose that $A$ is WOT$^\*$ sequentially compact; $A$ is bounded by the uniform boundedness principle. Then each sequence $(T\_n)$ in $A$ has a WOT$^\*$ convergent subsequence. Thus the image sequence in $C(M)$ has a pointwise convergent subsequence. By dominated convergence it is weakly convergent in $C(M)$, so $(T\_n)$ has a weakly convergent subsequence. Therefore $A$ is weakly compact and hence WOT$^\*$ compact.


The converse follows from the stated theorems of Kalton and Eberlein-Shmulyan.


