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Let $T:E\rightarrow F$ be a bounded linear map for E and F Banach spaces, and E reflexive. Let $B_E$ be the unitary closed ball in $E$. How would you argue that $T(B_E)$ is closed?

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If $E$ is reflexive, its closed unit ball is weakly compact, and $T$, being continuous from the weak topology to the weak topology, takes weakly compact sets to weakly compact sets.

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  • $\begingroup$ But how do I connect weakly compact with closed? $\endgroup$ Nov 28, 2012 at 23:17
  • $\begingroup$ why is $T$ continuous with respect to weak topologies? $\endgroup$
    – Norbert
    Nov 29, 2012 at 0:10
  • $\begingroup$ Because the weak topology is the weakest topology that makes every lineal function continuous. $\endgroup$ Nov 29, 2012 at 0:23
  • $\begingroup$ ... every bounded linear functional, that is. And if $\varphi$ is a bounded linear functional on $F$, $\varphi \circ T$ is a bounded linear functional on $E$. $\endgroup$ Nov 29, 2012 at 3:11
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    $\begingroup$ Weakly compact implies weakly closed, and weakly closed implies closed. $\endgroup$ Nov 29, 2012 at 21:12

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