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.
Best Answer
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).