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If $E$ and $F$ are two normed vector spaces, $f:E\rightarrow F$ is a linear-continuous-bijective function. Then naturally I would think that $f^{-1}$ is also linear-continuous-bijective. But the textbook says that $f^{-1}$ can be non-continuous. I can't figure out why.

Can anyone raise a simple example?

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Same strategy as in the previous: write down a bounded invertible operator whose inverse isn't bounded. –  Qiaochu Yuan May 27 '12 at 5:16
    
My bad. Asking without thinking. So the critical problem I have here is my intuition is limited within the finite dimensional spaces, while things are much different in the infinite dimensional spaces. –  Voldemort May 27 '12 at 5:26
    
The construction is very similar to the construction I used in your other question. Just think about it for a bit. I would recommend using the same normed spaces again or perhaps a slight modification. –  Qiaochu Yuan May 27 '12 at 5:29
    
Thank you. I managed to construct an example. –  Voldemort May 27 '12 at 5:31

1 Answer 1

up vote 1 down vote accepted

Let $E=F=c_{00}$ be the subspace of $\ell_\infty$ consisting of sequences with finite support. Consider the linear map $f:c_{00}\to c_{00}$ which sends $e_n\to \frac 1 n e_n$ where $\{e_n\}$ denotes the canonical basis.

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