# Weak sequential compactness on the space of compact operators

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)

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.