All constructions of a discontinuous linear map from an Hilbert Space, that i have seen, rely on using the Axiom of Choice. A lot of theorems that heavily rely on AC are equivalent to AC itself so i asked myself if that's also the case here. But how would one even prove that and what general properties of a dicontinuous linear map would be useful to prove (or disprove) that statement?
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1$\begingroup$ en.wikipedia.org/wiki/… I some models of ZF, every linear map on a Banach space is continuous. $\endgroup$– Anne BauvalCommented Jun 6 at 18:57
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3$\begingroup$ @AnneBauval The OP asks if the existence of a discontinuous linear map over a Hilbert space is equivalent to AC. $\endgroup$– Hanul JeonCommented Jun 6 at 19:15
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No, not even close.
Hahn–Banach is equivalent to the statement that every Banach space has a discontinuous linear functional. As every Hilbert space is a Banach space, this is already a seemingly stronger statement.
And seeing how the Hahn–Banach theorem is even weaker than the Boolean Prime Ideal theorem, which itself is weaker than the Axiom of Choice, the answer to your question is negative.
answered Jun 6 at 22:19
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$\begingroup$ I do not quite understand how Hahn Banach implies the existence of a discontinuous linear functional on every Banach space. $\endgroup$ Commented Jun 7 at 14:42
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$\begingroup$ Futhermore i just found this: homepages.math.uic.edu/~rosendal/PapersWebsite/… , which claims the opposite. $\endgroup$ Commented Jun 7 at 14:52
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$\begingroup$ I couldn't find the word "Hahn" in that document. So I'm not sure what you mean by "claims the opposite". $\endgroup$– Asaf Karagila ♦Commented Jun 7 at 15:40
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$\begingroup$ "On another note, by results of R.M. Solovay [27] and S. Shelah [25], it is known to be consistent with ZF that all sets of reals are Baire measurable, which implies that all subsets of Polish groups are Baire measurable. Therefore, by Pettis’ Theorem, it is consistent with ZF that all homomorphisms between Polish groups are continuous. Thus, in order to produce discontinuous homomorphisms the axiom of choice must intervene in some fashion, e.g., via the existence of ultrafilters, Hamel bases, etc." It's on the 4th page of the pdf. $\endgroup$ Commented Jun 7 at 18:45
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1$\begingroup$ Yes, AC is needed, but Hahn–Banach suffices, which is significantly weaker than AC itself. $\endgroup$– Asaf Karagila ♦Commented Jun 7 at 20:49