If a map $C:X\rightarrow U$ maps every weakly convergent sequence into strongly convergent

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

Solution 1:

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 $X$).

Solution 2:

If $C$ is not linear the correct definition is: $C$ is compact if $C(A)$ is pre compact whenever $A$ is bounded. If $C$ is linear we get the usual definition by homogeneity.

Now, if $X$ is reflexive the answer is true. To see this, let $v_n\in C(A)$ where $A\subset X$ is a bounded set and note tat $v_n=C(u_n)$, with $u_n \in A$. Because $X$ is reflexive we can extract a subsequence of $u_n$ (not relabeled), such that, $u_n\rightarrow u$ weakly in $X$. Hence, by using the hypothesism we get that $C(u_n)\rightarrow C(u)$ strongly in $\overline{T(A)}$.