Let M be a dense infinite dimensional subspace in a Hilbert space H. Is there some bounded linear one-to-one operator T on H such that Ran(T) $\subseteq$ M?
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3$\begingroup$ No. Take the linear subspace of finitely supported vectors in $\ell^2$. I wonder what is the example of the person who downvoted... $\endgroup$– fedjaJun 5, 2015 at 21:31
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1$\begingroup$ Same as yours, fedja. $\endgroup$– Bill JohnsonJun 5, 2015 at 23:15
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$\begingroup$ Is anyone easily able to characterize the possible sorts of subspaces that might be these images? Without thinking too much about it, the thought insinuates itself on me that this question might encode a great many profound (and not presently answerable) questions. $\endgroup$– paul garrettJun 5, 2015 at 23:35
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$\begingroup$ @paul garrett We can certainly discuss that but I'd like to see the looming closure relieved first. One way to do it is to ask a new question in the form that would provoke such a discussion. Do you want to do it, Paul? $\endgroup$– fedjaJun 6, 2015 at 2:38
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3$\begingroup$ I wish people had voted to migrate this to Math.SE where it could get a real answer. But anyway, I think it is false even if we drop the word "bounded"; by Baire category, the Hamel dimension of an infinite-dimensional Hilbert space is $> \aleph_0$ (and in fact it can be shown it is at least $\mathfrak{c}$), hence the same is true of any injective linear image. But a separable Hilbert space has dense subspaces of Hamel dimension $\aleph_0$. $\endgroup$– Nate EldredgeJun 6, 2015 at 16:53
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