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For $n\in\mathbb N$, let \begin{align*} B^n\equiv&\;\{\mathbf x\in\mathbb R^n\,|\,\lVert \mathbf x\rVert\leq 1\}\text{ and}\\ S^{n-1}\equiv&\;\{\mathbf x\in\mathbb R^n\,|\,\lVert \mathbf x\rVert= 1\} \end{align*} denote the unit ball and the unit sphere, respectively, where $\lVert\cdot\rVert$ is the standard Euclidean norm.

Suppose that $F:B^n\to\mathbb R^n$ is a continuous function satisfying $F(-\mathbf x)=-F(\mathbf x)$ for every $\mathbf x\in S^{n-1}$. That is, $F|_{S^{n-1}}$ is an odd function.

 Claim: The function $F$ has a fixed point, that is, some $\mathbf x^*\in B^n$ satisfying $F(\mathbf x^*)=\mathbf x^*$.

(The case $n=1$ is an immediate consequence of the intermediate-value theorem.)

(Disclosure: I spent the better half of this afternoon trying to construct a counterexample for $n=2$, in vain.)


Note: The function $F$ does not necessarily map into $B^n$, so Brouwer’s fixed-point theorem doesn’t apply directly. That said, upon defining the projection function $\mathsf p:\mathbb R^n\to B^n$ as \begin{align*} \mathsf p(\mathbf x)\equiv\frac{1}{\max\{\lVert\mathbf x\rVert,1\}}\mathbf x\quad\text{for each $\mathbf x\in\mathbb R^n$}, \end{align*} the composite function $\mathsf p\circ F:B^n\to B^n$ is guaranteed to have a fixed point. Yet, it is unclear to me whether and how this observation helps prove that $F$ has a fixed point. (I would like to avoid using the Borsuk–Ulam theorem and heavy-duty algebraic-topology arguments if possible. Brouwer’s level is as high as I’d like to preferably reach.)

Any suggestions would be appreciated.

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Hopefully another answerer will have an idea about how to do this without a little algebraic topology machinery, but here is an initial solution.

You are correct that $F$ must have a fixed point. Proof by contradiction:

$F$ is defined on the compact set $B^n$, so the image is compact. Thus both $B^n$ and $F(B^n)$ are contained in some sphere $S$ (of dimension $n-1$) of large radius, centered at the origin.

Suppose $F$ has no fixed point. Then (using the same trick in the proof of the Brouwer f.p. theorem) for each $x\in B^n$, there is a unique ray from $x$ to $F(x)$, hitting $S$ in a unique point, and this point varies continuously with $x$. Thus we construct a continuous map $\Phi: B^n\rightarrow S$.

The condition that if $x\in S^{n-1}$ then $F(-x) = -F(x)$ means that the ray from $x$ to $F(x)$ negates the ray from $-x$ to $F(-x)$, and so their intersections with $S$ are antipodes. This means that the restriction of $\Phi$ to the boundary $S^{n-1}$ sends antipodes to antipodes.

Thus if $i:S^{n-1}\hookrightarrow B^n$ is the inclusion, the composed map $$S^{n-1}\xrightarrow{i} B^n\xrightarrow{\Phi} S$$ sends antipodes to antipodes.

By a theorem of Borsuk (which is supposed to be equivalent to the Borsuk-Ulam theorem, although I'm afraid I don't know why), a map from the sphere to the sphere that preserves antipodes must have odd degree. In particular, the degree of $\Phi\circ i$ is not zero. So the induced map on homology

$$H_{n-1}(S^{n-1})\xrightarrow{i_\star} H_{n-1}(B^n) \xrightarrow{\Phi_\star} H_{n-1}(S)$$

is not zero. But this contradicts the fact that $H_{n-1}(B^n) = 0$ (since $B^n$ is contractible).

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  • $\begingroup$ I see, thank you for the affirmative answer. Unfortunately, I know next to nothing about algebraic topology, so you lost me at homologies. An interesting question is whether the following result can be deduced from my claim: “There exists no continuous $g:B^n\to S^{n-1}$such that $g(-\mathbf x)=-g(\mathbf x)$ for every $\mathbf x\in S^{n-1}$.” This statement, too, is equivalent to Borsuk–Ulam, so if it turns out that it is strictly stronger than the fixed-point theorem under consideration, then there is some hope that an easier proof of the latter (hopefully based on Brouwer) is available. $\endgroup$ – triple_sec Sep 26 '18 at 3:41
  • $\begingroup$ Here is a possible approach. Is it true that every continuous odd function from $S^{n-1}$ to $S^{n-1}$ is a homeomorphism? Suppose it is true and it can be proved by elementary methods. Then, your proof can be finished in the following way: $$B^n\xrightarrow{\Phi}S\xrightarrow{h}S^{n-1}\xrightarrow{g}S^{n-1},$$ where $h$ is the “shrinking” homeomorphism from $S$ onto $S^{n-1}$, and $g$ is the inverse of $h\circ\Phi|_{S^{n-1}}$. This way, one obtains a retraction $g\circ h\circ \Phi$ from $B^n$ onto $S^{n-1}$, which I already know from Brouwer is impossible. $\endgroup$ – triple_sec Sep 26 '18 at 4:04
  • $\begingroup$ In reply to your first comment, what is clear to me is the other direction: your fixed-point claim in the OP can be deduced from the statement that there is no continuous $B^n\rightarrow S^{n-1}$ that has odd symmetry on the boundary. In fact, that's essentially what my argument is doing. If you believe this statement, you can skip the homology part of my arg. and just observe that I'm showing how, given hypothetical fixed-pt-free $F$, to construct a continuous map $B^n\rightarrow S$ with odd symmetry on the boundary, which you believe cannot exist. $\endgroup$ – Ben Blum-Smith Sep 26 '18 at 13:36
  • $\begingroup$ I don't see right away how to do the converse you asked about (i.e. deduce "no $B^n\rightarrow S^{n-1}$ with odd symmetry on the boundary" from your fixed-point claim), but it's plausible to me that this is possible. $\endgroup$ – Ben Blum-Smith Sep 26 '18 at 13:38
  • $\begingroup$ In response to the question in your second comment, the answer is no. You can get a counterexample just by perturbing the identity map locally in the neighborhood of a pair of antipodes til it is not injective in these two small regions. But actually there are odd maps that are not even homotopic to a homeomorphism! Example: identify $S^2$ with the extended complex plane (i.e. Riemann sphere), so $z$ and $-1/\overline z$ are antipodes. Then $z\mapsto z^2(1-z)/(1+z)$ is such a map. See math.univ-toulouse.fr/~buff/Preprints/Antipodal/Antipodal1.pdf $\endgroup$ – Ben Blum-Smith Sep 26 '18 at 13:48

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