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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

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  • $\begingroup$ Is $C$ a linear map? If not, your definition of compact is wrong. $\endgroup$ – Tomás Dec 5 '12 at 12:47
  • $\begingroup$ yes thanks, I changed it! $\endgroup$ – Johan Dec 5 '12 at 13:58
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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 $\overline{T(A)}$.

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    $\begingroup$ nice! why cant we extract this if $X$ is not reflective!? $\endgroup$ – Johan Dec 5 '12 at 13:56
  • $\begingroup$ If $X$ is not reflexive, you cant ensure the existence of weakly convergent subsequence. In fact you can guarantee the existence of a weakly star convergent subsequence and if $X$ is reflexive, the weak star topology and the weak topology coincide, hence, you can conclude weak convergence. Take a look in David Mitra example, note that $\ell_1$ is not reflexive. $\endgroup$ – Tomás Dec 5 '12 at 14:22
  • $\begingroup$ What is $\overline{T(A)}$ here? Did you mean $\overline{C(A)}$? $\endgroup$ – Ryker Apr 11 '16 at 2:28
  • $\begingroup$ Yes @Ryker, you are right. $\endgroup$ – Tomás Apr 11 '16 at 21:16
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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 $X$).

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  • $\begingroup$ Thanks! great counterexample! $\endgroup$ – Johan Dec 5 '12 at 13:57

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