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Let $(X,|.|)$ be a Banach space. $A\in B(X)$ a bounded injective operator. Then we can define another norm on $X$ by $$|x|_A=|Ax|.$$ Since we have $$|x|_A\leq |A||x|$$ Then by the result of continuity of the inverse, there's a constant $c>0$ such that $$|x|\leq c|x|_A=c|Ax|$$ But the last inequality means that $A$ cannot be compact. This means every injective bounded operator is not compact which is not true because there's lot of counter examples. So i don't know where I am wrong in my reasoning.

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    $\begingroup$ Your "result of continuity of the inverse" requires that the range of $A$ be closed. $\endgroup$ – David Mitra May 21 '14 at 11:15
  • $\begingroup$ @David I just saw that $(X,|.|_A)$ is actually not a Banach space, so I can't apply this result. Which as you said needs $R(A)$ to be closed. This maybe provides a counter example of applying the open map theorem with one of domain/codomain not Banach. $\endgroup$ – user165633 May 21 '14 at 11:23
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As David Mitra mentioned, $(X,|.|_A)$ is not Banach, so we can't apply the result of the boundedness of the inverse operator. This needs $A$ to have a closed range.

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