# A property of even continuous functions on the sphere

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.

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

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 $$3(n+\frac{1}{3})\neq0$$.