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Set theorists typically prove the existence of Banach limits (EBL) using the Ultrafilter Theorem or, its equivalent, the Boolean Prime Ideal Theorem (BPI). Analysts, on the other hand, typically prove the existence of Banach limits using the Hahn-Banach Theorem (HB). It is known that ZF + BPI implies HB, but ZF+ HB does not imply BPI. This raises the question if ZF+ HB implies EBL or do the proofs employed by analyst’s (of necessity) make use (tacit or otherwise) of a consequence of the Axiom of Choice in addition to HB?

Edit: The proof given by J. B. Conway, in A Course in Functional Analysis, Graduate Texts in Mathematics. Vol. 96. New York: Springer, 1994, does not appear to make an overt use of a consequence of the Axiom of Choice other than HB.

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    $\begingroup$ I believe it really is just a straightforward application of the Hahn–Banach theorem, with no further choice: you extend the linear functional ${\lim}\colon c\to\mathbb R$ to a linear functional $\ell_\infty\to\mathbb R$ bounded by the $\ell_\infty$ norm. $\endgroup$
    – Emil Jeřábek
    Commented Apr 26, 2022 at 12:58
  • $\begingroup$ @ Emil Jeřábek. Thanks! The proof offered by J.B. Conway referred to in my edit certainly seems to accord with your assertion. $\endgroup$
    – Philip Ehrlich
    Commented Apr 26, 2022 at 13:02
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    $\begingroup$ I forgot about the shift-invariance condition. But this is free for all: it suffices to compose the functional with the arithmetic mean operator $A\colon\ell_\infty\to\ell_\infty$ mapping $(a_n)_{n\in\omega}$ to $(\frac1n\sum_{i<n}a_i)_{n\in\omega}$, as $A(a)-A(S(a))$ has (ordinary) limit $0$ for any $a\in\ell_\infty$. $\endgroup$
    – Emil Jeřábek
    Commented Apr 26, 2022 at 13:07
  • $\begingroup$ @ Emil Jeřábek. Yes, but I take it this does not impact the claim that ZF+ HB implies EBL--right? $\endgroup$
    – Philip Ehrlich
    Commented Apr 26, 2022 at 13:11
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    $\begingroup$ Yes, the extra argument is just in ZF. $\endgroup$
    – Emil Jeřábek
    Commented Apr 26, 2022 at 13:16

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