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)


1 Answer 1


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.


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