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This question is inspired by On moments of inertia of planar and 3D convex bodies. Let $f:{\mathbb R}^3\setminus\{0\}\to{\mathbb R}$ be an even homogeneous ($f(kx)=f(x)$ for all real $k\neq 0$) continuous function. Can one always find an orthogonal basis $(x_1,x_2,x_3)$ such that $f(x_1)=f(x_2)=f(x_3)$ ?

Remark. Even without the condition that $f$ is even, I do not know a counterexample. This condition comes from the application that I mention in the first sentence.

Remark 2. Thanks to @Aleksei Kulikov, who found the reference: this was a problem posed by Rademacher and solved by Kakutani in 1942 (Annals of math., 43 (1942) 739-741). Most interestingly, the reviewer in MR0007267 wrote that the corresponding problem in $R^n, n\geq 4$ was unsolved in 1942. I found that the $n$-diensonal generalization was obtained by H. Yamabe and Z. Yujobô, Osaka Math. J. 2 (1950), 19–22.

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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. $\endgroup$ Dec 4, 2022 at 18:19
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    $\begingroup$ Is there some easy counterexample if we don't require $f$ to be even? $\endgroup$
    – Saúl RM
    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). $\endgroup$ Dec 4, 2022 at 20:39
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    $\begingroup$ @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... $\endgroup$ Dec 4, 2022 at 20:54
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    $\begingroup$ 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. $\endgroup$ Dec 5, 2022 at 14:37

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It seems there aren't any counterexamples, even if $f$ is homogeneous but not even.

If there is some counterexample $f$ to the question, and letting $X=\mathbb{R}^3\setminus\{x_1=x_2=x_3\}$, we can consider the map $F:SO(3)\to X$, given by $F(u,v,w)=(f(u),f(v),f(w))$ (we will sometimes represent elements $r\in SO(3)$ by $(r(e_1),r(e_2),r(e_3))$, $(e_i)_{i=1}^3$ being the usual basis of $\mathbb{R}^3$).

As $\pi_1(SO(3))\cong\mathbb{Z}_2$ and $\pi_1(X)\cong\mathbb{Z}$, we will have $F_*(\gamma)=0\;\forall\gamma\in\pi_1(SO(3))$.

Now consider the loop $\gamma:t\mapsto\gamma_t=r_{2\pi t}(v)$ given by rotations of angle $2\pi t$ around $v:=(1,1,1)$. Note that $\gamma_{\frac{1}{3}}(e_1)=e_2,\gamma_{\frac{1}{3}}(e_2)=e_3$ and $\gamma_{\frac{1}{3}}(e_3)=e_1$. So if $\gamma_t=(u,v,w)$, then $\gamma_{t+\frac{1}{3}}=(\gamma_{t+\frac{1}{3}}(e_1),\gamma_{t+\frac{1}{3}}(e_2),\gamma_{t+\frac{1}{3}}(e_3))=(\gamma_{t}(e_2),\gamma_{t}(e_3),\gamma_{t}(e_1))=(v,w,u)$.

So the path $\alpha=F_*\gamma\in\pi_1(X)$ satisfies that if $\alpha(t)=(x_1,x_2,x_3)$, then $\alpha(t+\frac{1}{3})=(x_2,x_3,x_1)$. From this we will deduce that $\alpha$ is not trivial, which is a contradiction. To deduce that $\alpha$ is not trivial, we can first deformation retract $X$ to a circumference perpendicular to the line $\{x_1=x_2=x_3\}$, so that identifying this circumference with $\mathbb{S}^1$, the rotation $(x_1,x_2,x_3)\mapsto(x_2,x_3,x_1)$ becomes the rotation $r_{\frac{2\pi}{3}}$ of $\mathbb{S}^1$.

So we have to prove that any loop $\beta:[0,1]\to\mathbb{S}^1$ satisfying $\beta(t+\frac{1}{3})=e^{\frac{2\pi i}{3}}\beta(t)$ is not trivial. This is true because as $\beta(\frac{1}{3})=e^{\frac{2\pi i}{3}}$, the path $\beta|_{[0,\frac{1}{3}]}$ rotates a total angle of $2\pi(n+\frac{1}{3})$ for some $n\in\mathbb{Z}$, so the winding number of $\beta$ is $3(n+\frac{1}{3})\neq0$.

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