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This seems like a fundamental result but I can not solve it of find an solution: Let $M:X\rightarrow U$ be a bounded linear map between Banach spaces. Show that if the range of M is a set of second category of U; then the range is all of U.

Johan
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    This is very confusing to me. Your statement is not true and the title does not seem to involve the question at all. For a counterexample consider something like the diagonal operator $e_n \mapsto e_n/n$ on $\ell^1(\mathbb N)$. – JSchlather Nov 29 '12 at 17:23
  • @JacobSchlather, why don't you post this comment as answer? – Norbert Nov 29 '12 at 17:43
  • Sorry I wrote it wrong, it is corrected now. – Johan Nov 29 '12 at 17:44
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    @Norbert, I usually post as a comment when I think someone meant to ask something different and I'm unsure if I'll be able to answer their actual question. – JSchlather Nov 29 '12 at 17:56
  • The title doesn't seem to be at all related to the question. – Chris Eagle Nov 29 '12 at 18:28
  • I think the standard proof of the Open Mapping Theorem, where the hypothesis that "$M$ is onto" is replaced by "$\text{Ran}(M)$ is of second category" can be slightly altered to give your result. In fact, more is true: If $X$ is a Banach space, $Y$ is a normed space, $M:X\rightarrow Y$ is continuous, linear, and $M(X)$ is of second category, then: 1) $M$ is onto, 2) $M$ is an open mapping, and 3) $Y$ is complete. This (ever further generalized) is Theorem 2.11 in Walter Rudin's Functional Analysis. – David Mitra Nov 29 '12 at 18:35
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    Should the title say "second category" instead of "second countable"? – Nate Eldredge Nov 29 '12 at 18:39

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Assume that the range is of second category but the range of $M$ is not $U$. As $M(X)$ is not closed (otherwise $M(X)$ would be of first category, as it's a strict subspace), we can apply this result to $Y=\overline{M(X)}$.

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Davide Giraudo
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