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The title says it all. Given an arbitrary infinite dimensional normed vector space $X$, can you show there exists a discontinuous (perhaps unbounded) linear functional on $X$ without resorting to Axiom of Choice/Hamel basis techniques? I know similar questions have been asked an answered with these methods, but I know they are not necessary if the space is not Banach.

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  • $\begingroup$ Your question is unclear. Knowing nothing about $X$ it need not be possible even with the axiom of choice to present a discontinuous (which means the same as being unbounded (not perhaps)) linear functional on $X$. In particular, all linear functionals of finite dimensional real (or complex) vector spaces are continuous (i.e., are bounded). $\endgroup$ – Ittay Weiss Sep 20 '15 at 22:20
  • $\begingroup$ I wasn't asking for a presentation, just an existence. EDIT: Sorry, I forgot to say infinite dimensional! $\endgroup$ – Toeplitz Sep 20 '15 at 22:21
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    $\begingroup$ It is probably compatible with ZF that all linear functionals are continuous. $\endgroup$ – Mariano Suárez-Álvarez Sep 20 '15 at 22:24
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It is consistent with $ZF$ that every linear function from $\mathbb{R}$ (viewed as a vector space over $\mathbb{Q}$) to $\mathbb{R}$ is continuous; see https://mathoverflow.net/questions/57426/are-there-any-non-linear-solutions-of-cauchys-equation-fxy-fxfy-wit. So the answer to your question is that some amount of choice is required, even for not-too-nasty vector spaces.

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  • $\begingroup$ Well, then, is it possible to do such a thing without the notion of a Hamel basis? $\endgroup$ – Toeplitz Sep 21 '15 at 0:51
  • $\begingroup$ @Lusin What do you mean "do such a thing without the notion of a Hamel basis?" Are you asking whether a proof that such functions exist, which doesn't use the language of Hamel bases, is known? $\endgroup$ – Noah Schweber Sep 21 '15 at 0:53
  • $\begingroup$ Yes. I am looking for an answer to the original problem which avoids the use of Hamel bases. $\endgroup$ – Toeplitz Sep 21 '15 at 0:53

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