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Let $B$ be an Infinite dimensional Banach Space and $T:B\to B$ be an continuous operator such that $T(B)=B$ and $T(x)=0\Rightarrow x=0$

which of the following is correct?

  1. $T$ maps bounded sets into compact set

  2. $T^{-1}$ maps bounded sets into compact set

  3. $T^{-1}$ maps bounded sets into bounded set

  4. $T$ maps compact sets into open set I have no idea how to do it.

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Isn't this the bounded inverse theorem (en.wikipedia.org/wiki/Bounded_inverse_theorem), and hence the 3rd option? –  Jan Keersmaekers Sep 12 '13 at 12:48
    
Everything is always mapped into open sets. Should 4. say "onto open sets"? –  Jonas Meyer Sep 12 '13 at 17:53
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1 Answer 1

up vote 4 down vote accepted

$1.$ and $2.$ are not satisfied by $Tx=x$.

$3.$ $T^{-1}$ is bounded (as a consequence of the closed graph theorem).

$4.$ Take $K:=\{0\}$.

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Davide could you tell me how $T^{-1}$ is bounded? –  miosaki Sep 13 '13 at 5:12
    
What did your attempt with closed graph theorem give? –  Davide Giraudo Sep 13 '13 at 9:32
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