Let $E$ be a normed (real) space which is not complete. Is it always possible to find $f$ a continuous bijective linear function from $E$ to $E$ such that $f^{-1}$ is not continuous?
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asked Jul 8, 2012 at 8:55
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No. If $E$ has codimension less than the continuum in a Banach space then such an $f$ must be open. This was proved by Saxon and Levin; see the proposition on page 95 of
which is
Saxon, Stephen; Levin, Mark, Every countable-codimensional subspace of a barrelled space is barrelled. Proc. Amer. Math. Soc. 29 1971 91–96.
So for an example, take any discontinuous linear functional on a Banach space and let $E$ be its kernel.
answered Jul 8, 2012 at 17:44
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$\begingroup$ Thanks Bill. What would you recommend as books to go between "initial functional analysis" (Banach isomorphism theorem, Banach-Steinhaus theorem...) and the research articles you mention? I have to bridge the gap! $\endgroup$ Jul 8, 2012 at 19:52
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$\begingroup$ Sorry, I don't. This is TVS theory rather than Banach space theory. I would look at the books referenced in the Levin-Saxon paper; i.e., Edwards' book and Bourbaki's book on TVS (v. 5). Probably one of them contains the result that finite codimensional subspaces of Banach spaces are barrelled. IIRC, Edwards' book is reader friendly, while Bourbaki's has the advantage of being in French. $\endgroup$ Jul 8, 2012 at 20:05