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Let $X$ and $Y$ be some banach spaces. Then the bounded linear operator $T:X\longrightarrow Y$ has bounded $f^{-1}$ if and only if it is bijective (this is the so called Banach open mapping theorem). Is it true that if the given normed space $X$ is not banach then there exists one-to-one bounded linear operator $T:X \longrightarrow X$ whose $T^{-1}$ is not bounded?
What is the easiest example of bounded $T:X\longrightarrow X$ whose $T^{-1}$ is not bounded?

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  • $\begingroup$ See this. $\endgroup$ – David Mitra Mar 7 '14 at 18:19
  • $\begingroup$ If you really want $T:X\rightarrow X$, here's an example. Take $X$ to be the space of infinite sequences with finite support with the sup norm. Map the $n$'th standard unit vector $e_n$ to $e_n/n$ and extend. This gives a bijective linear contraction whose inverse is not bounded. $\endgroup$ – David Mitra Mar 7 '14 at 19:21

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