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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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But how do I connect weakly compact with closed? – Rojas Azules Nov 28 '12 at 23:17
why is $T$ continuous with respect to weak topologies? – Norbert Nov 29 '12 at 0:10
Because the weak topology is the weakest topology that makes every lineal function continuous. – Rojas Azules Nov 29 '12 at 0:23
... 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 $E$. – Robert Israel Nov 29 '12 at 3:11
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Weakly compact implies weakly closed, and weakly closed implies closed. – Robert Israel Nov 29 '12 at 21:12
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