https://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem#Relation_to_axiom_of_choice (current revision)
As mentioned earlier, the axiom of choice implies the Hahn–Banach theorem. The converse is not true. One way to see that is by noting that the ultrafilter lemma, which is strictly weaker than the axiom of choice, can be used to show the Hahn–Banach theorem, although the converse is not the case. The Hahn–Banach theorem can in fact be proved using even weaker hypotheses than the ultrafilter lemma.[4] For separable Banach spaces, Brown and Simpson proved that the Hahn–Banach theorem follows from WKL0, a weak subsystem of second-order arithmetic.[5]
