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In this post, it is said that there are no constructive examples of everywhere-defined unbounded operators between Banach spaces; every example furnished must use the axiom of choice. This seems like a rather surprising claim. Can someone provide proof of this fact or point me to a reference?

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  • $\begingroup$ Note that e.g. infinite dimensional separable Banach spaces $X, Y $ even not isomorphic still have the same Hamel dimension, so you know there exists a bijective unbounded operator $X\to Y$ $\endgroup$
    – Pietro Majer
    Commented Jun 15 at 18:37
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    $\begingroup$ Assuming that the Principle of Dependent Choice and that every set of reals has the Baire Property (or Lebesgue measurable), every group homomorphism between completely metrisable groups is continuous. Therefore, in that case, any linear map between Banach spaces is continuous, and therefore is bounded. $\endgroup$
    – Asaf Karagila ♦
    Commented Jun 15 at 18:44

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