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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.

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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
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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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That was a long one! Is there no easier way? –  Johan Nov 30 '12 at 9:49
    
I don't know, but what we have to show is strongly related to what is done in the link. –  Davide Giraudo Nov 30 '12 at 12:46
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