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Dec 2 at 7:10 history edited Jochen Glueck CC BY-SA 4.0
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Sep 4 at 10:13 comment added Jochen Glueck @euleroid: Thanks, you're right that proof in the paper, if it were correct, would actually show a slightly weaker claim than I initially wrote. I've edited my answer accordingly. Please note, though, that I'm not saying that the paper explicitly make this claim. I'm just saying that the proof would imply this claim if the proof were correct. Note that the proof does not really use that $T$ has no invariant subspace; it just uses that (1) is true for at least one vector $x$ and for the vector $Tx$.
Sep 4 at 10:04 history edited Jochen Glueck CC BY-SA 4.0
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Sep 4 at 9:39 comment added euleroid I agree with you that $f$ may not be well-defined, but in terms of a conceptual problem, I don't think that the paper makes the claim you say, since if $T$ is an operator with no invariant subspaces that is enough to conclude that every orbit is dense, since if there was a non-dense orbit it would be an invariant subspace.
Sep 4 at 6:58 history edited Jochen Glueck CC BY-SA 4.0
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Sep 4 at 6:54 comment added Jochen Glueck Now, the article names Palle Jorgensen as its editor. What might be going on there?
Sep 4 at 6:46 history answered Jochen Glueck CC BY-SA 4.0