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.
2026-03-30 16:21:46.1774887706
Where am I wrong ??
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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.