# Generalization of Borsuk-Ulam to arbitrary ratio

Let $g: S^n \to R^n$ be a continuous odd function (i.e. $g(-x)=-g(x)$ for all $x$). The Borsuk-Ulam theorem implies that $g$ has a zero, i.e. there is an $x$ such that $g(x)=(0,0,...,0)$.

Suppose $g$ is (1,1,...,1) on the positive orthant (i.e. when all its $n$ arguments are non-negative) and (-1,-1,...,-1) on the negative orthant. Is this true that for every constant $r\in [-1,1]$, there is an $x$ such that $g(x)=(r,r,...,r)$?

For $n=1$, this is obviously true by the intermediate value theorem. Under what conditions is it true for $n>1$?

The question seems closely related to the Poincare-Miranda theorem, which is a generalization of the IVT to multi-dimensional cubes, but so far I haven't found the connection.

• What if you take $f \colon S^1 \to \mathbb{R}$ given by $f(x, \, y)=x^2$? This is a continuous, non-negative function with two zeroes on $S^1$, such that one has $f(-x, \, -y) = f(x, \, y)$ for every point, right? – Francesco Polizzi May 18 '15 at 10:11
• @FrancescoPolizzi Thanks, I had to add an additional condition. – Erel Segal-Halevi May 18 '15 at 11:23

I believe I have a valid counterexample for $n > 1$, unfortunately I don't have the expertise to be certain. I would glad if someone would expand or refute the following.

Take $n=2$ to start. Now we are going to create a pair of functions $f_1(x),f_2(x)$ with $g(x)=(f_1,f_2)$ which don't satisfy the property. First take $f=f_1'=f_2'$ to be a simple bipolar function such that $f$ is $0$ on a great circle and has a maximum at $x_m = (1,1)$ and minimum at $(-1,-1)$ s.t. $g(x_m) = M < 1$.

Now we modify $f_1'$ into $f_1$ to have a continuous bump around $x_m$ of radius $r_1$ and value $1$, that is, $g(x) \approx (f_1',*)$ when $|x_m \cdot x| \le 1-h(r_1)$ and $g(x) \approx (1,*)$ when $|x_m \cdot x| \gt 1-h(r_1)$, and accordingly we modify $f_2$ to have a bump around $x_m$ of radius $r_2$ and value $1$, that is, $g(x) \approx (*,f_2')$ when $|x_m \cdot x| \le 1-h(r_2)$ and $g(x) \approx (*,1)$ when $|x_m \cdot x| \gt 1-h(r_2)$. So that we have $g=(f_1,f_2)$.

Now if $h(r_1) \ll h(r_2) \ll 1$, then $f_1$ assumes almost all the values in $[M,1]$ near a circle of radius $r_1$, while $f_2$ assumes almost all values in $[M,1]$ near a circle of radius $r_2$. Clearly, those concentric circles don't intercept. We conclude there exists $y \in [M,1]$ such that $g(x)=(y,y)$ has no solution.

I could generalize this, if I'm not missing something.

• What do you mean by the "*" symbol? – Erel Segal-Halevi May 24 '15 at 18:48
• Maybe you mean something like this: tube.geogebra.org/student/m1236259 Suppose f is 1 on the green face and -1 on the red face, and its value elsewhere depends only on its distance from the green face (e.g. it has a single value on the entire yellow line). Apparently there can be many functions that satisfy this condition, and they don't have to coincide anywhere in the blue region. – Erel Segal-Halevi May 24 '15 at 19:04
• Ah yes, something like that. The idea was to make the lines $f_1=k$ and $f_2=k$ parallel, by making both $f_1$ and $f_2$ a function only of the distance to two faces as you describe, or more simply a positive function of $x_m \cdot x$ (where $\cdot$ is the dot product). Indeed I think something like this gives a generalization, and a way to built $g$ such that $g(x)=(y,...,y) /iff x=x_m, x \cdot x_m = 0 or x=-x_m$. To that end, just choose two positive functions s.t. $f_1=h_1(x_m \dot x)$, $f_1=h_1(x_m \dot x)$, $h_1(y)=-h_1(y)$,$h_1(1)=-h(-1)=1=h_2(1)=-h_2(-1)$ that only intersect at 0,1,-1. – Real May 24 '15 at 21:12