My problem:
Suppose $D \subset \mathbb{R}^{m+1}$ is the unitary open ball. Let $f : \overline{D} \to \mathbb{R}^{m+1}$ be a continuous function. Suppose also that $\langle x,f(x) \rangle <1$ for all $x \in \mathbb{S}^{m}$. I would like to prove that $f$ has a fixed point.
My attempt:
I tried defining $f_n=\frac{f}{n}$ where $n \geq N=\sup\limits_{x \in \overline{D}} \|f(x)\|$ and taking $x_n$ the fixed point of $f_n$. Taking a subsequence we may assume $x_n \to x$.
Thus from $n \langle x_n,x_n \rangle <1$ we get $x_n \to 0$. But I cannot conclude.
Hint Consider $g\colon\overline{D}\to\overline{D}$ defined by $$ g(x):=\begin{cases} f(x) & f(x)\in\overline{D}\\ \frac{f(x)}{\lvert f(x)\rvert} & f(x)\notin\overline{D} \end{cases} $$