Every continuous function $f$ on the unit sphere (shell) has a point $p$ such that $f(p)=f(-p)$

Question

The Question:

Define the subset of $\Bbb R^3$

$$S=\{ (x,y,z) \in \Bbb R^3: x^2+y^2+z^2=1 \}$$

Suppose that $f:S \rightarrow \Bbb R$ is a continuous function. Prove that there exists $p \in S$ such that $f(p)=f(-p)$.

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My Attempt:

I don't know where to start really. I feel that the following facts might be useful:

• Intermediate Value Theorem
• $f:S \rightarrow \Bbb R$ is continuous and $S$ is connected $\implies$ $f(S)$ is connected
• Path connected $\implies$ connected

Any hints as to how I should tackle this problem?

Answer 1

This is a special case of the Borsuk-Ulam theorem. That states that if $f:S^n\to\Bbb R^n$ is continuous, then there is a $p$ with $f(p)=f(-p)$.

In our case just look for a point on $S^1=\{(x,y,0):x^2+y^2=1\}\subset S^2$ with $f(p)=f(-p)$. In this case, Borsuk-Ulam is quite elementary. Just consider $g(t)=f(\cos t,\sin t,0)-f(-\cos t,-\sin t,0)$. All you need is the Intermediate Value Theorem to ensure that $g$ has a zero.

Answer 2

If f is continuous, then $T: p\mapsto f(p) - f(-p)$ is continuous. By symmetry $T(p) = - T(-p)$. Since images of connected sets under continuous functions are connected this yields that the image of T is a symmetric interval containing $0$.