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?
$T(B_E)$ is closed for $T$ a bounded linear map
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functional-analysis
1 Answers
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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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0But how do I connect weakly compact with closed? – 2012-11-28
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0why is $T$ continuous with respect to weak topologies? – 2012-11-29
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0Because the weak topology is the weakest topology that makes every lineal function continuous. – 2012-11-29
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0... 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 $E$. – 2012-11-29
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0@RobertIsrael you are right my bad – 2012-11-29
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0I'm really sorry but I still don't see how is that weakly compact implies closed. – 2012-11-29
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2Weakly compact implies weakly closed, and weakly closed implies closed. – 2012-11-29