# 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? May 18 '15 at 10:11

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? 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. 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