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  • $\begingroup$ Since $f$ is even, it's tempting to identify antipodal points. For functions on $S^1$, this shows that the corresponding statement is correct and in fact the same as the (trivial, in $d=1$) Borsuk-Ulam theorem. In your actual setting, for $d=2$, we obtain the projective plane instead of the sphere, and the orthogonality condition does not seem to have a good description there. $\endgroup$
    – Christian Remling
    Commented Dec 4, 2022 at 18:19
  • 2
    $\begingroup$ Is there some easy counterexample if we don't require $f$ to be even? $\endgroup$
    – Saúl RM
    Commented Dec 4, 2022 at 19:19
  • $\begingroup$ @Saul RM: no even if one drops the evenness condition, I do not know a counterexample. Evenness condition come from the application that I had in mind (see the link in the question). $\endgroup$
    – Alexandre Eremenko
    Commented Dec 4, 2022 at 20:39
  • 1
    $\begingroup$ @Christian Remling: you are right; this is about continuous functions on the real projective plane. Standard spherical Riemannian metric (in which the diameter of the sphere is $\pi$) descends to the projective plane, and the diameter becomes $\pi/2$. So the question is: can you find three "diametrally opposite" points on the projective plane, such that... $\endgroup$
    – Alexandre Eremenko
    Commented Dec 4, 2022 at 20:54
  • 5
    $\begingroup$ This is actually a famous result of Kakutani (A Proof That there Exists a Circumscribing Cube Around Any Bounded Closed Convex Set in $\mathbb{R}^3$) and the proof is more or less the same as in the accepted answer. $\endgroup$
    – Aleksei Kulikov
    Commented Dec 5, 2022 at 14:37