A Linear map between Banach spaces $C:X\rightarrow U$ is compact if it maps if the closure of the image of the unit ball is precompact in U.
If a map $C:X\rightarrow U$ maps every weakly convergent sequence into strongly convergent
can we say that the map is compact?
The converse can be seen here:
Compact operator maps weakly convergent sequences into strongly convergent sequences
Best Answer
No. For example, in $\ell_1$ ($=\ell_1(\Bbb N)$) every weakly convergent sequence is norm convergent. Thus the identity operator, $I$, on $\ell_1$ maps weakly convergent sequences to norm convergent sequences. But $I$ is not a compact operator.
Incidentally, operators that map weakly convergent sequences to norm convergent sequences are called completely continuous. If $X$ is a reflexive space, then every completely continuous operator on $X$ to a Banach space $Y$ is compact (as a result that the closed unit ball of $X$ is weakly compact for reflexive $X$).