Fixed Point of continous map on a closed ball to R^n

Question

For a closed ball $\overline{B_R(0)}\subset \mathbb{R}^n$ with $R>0$ we have $f\in C^0(\overline{B_R(0)},\mathbb{R}^n)$ with $\langle x,f(x) \rangle\leq \langle x,x \rangle$ for $x \in \partial \overline{B_R(0)}$. Show that $f$ has fixed point.

My guess (I'm very sceptical): $f$ continuous implies that Im$(f)$ is compact. \begin{align} \text{Im}(f)\subset \overline{B_{R_i}(0)} \;\; \text{for} \;\; R_i\geq\!\!\max_{x\in \overline{B_R(0)}}\! ||f(x)||. \end{align} If $R>R_i$ I'm done, since I can use Brouwer fixed point theorem. For $R<R_i$ define $g_i: \mathbb{R}^n\rightarrow \mathbb{R}^n, x\mapsto \frac{R}{R_i} x$, so that $g_i\circ f(\overline{B_{R_i}(0)})\subset \overline{B_{R_i}(0)}$ and by Brouwers theorem there exists $x_0$ so that $g_i\circ f(x_0)=x_0$, hence there exists $x_0$ so that $f(x_0)=\frac{R_i}{R}x_0$ is true. But since $R_i$ can chosen arbitrary large, $x_0$ has to be $0$.

Does it work?

Answer 1

Hint: Try using the following well-known result in fixed point theory:

(Leray-Schauder Nonlinear Alternative): Let $E$ be a Banach space and $C \subset E$ a closed convex set. Let $p \in \text{int}(C)$ and $U\subset C$ be an open set containing $p$. If $f: \overline{U} \longrightarrow C$ is continuous and compat (that is, $\text{Im}(f)$ is contained in a compact of $C$), then either:

$(1)$ $f$ has a fixed point in $\overline{U}$; or

$(2)$ There is some $u \in \partial U \subset C$ and some $\lambda \in (0,1)$ such that

$$u=\lambda \cdot f(u) + (1-\lambda)\cdot p$$

What should $E$, $C$, $p$ and $U$ be in your example? Is $f$ compact (duh)? After answering these questions, all you have to do is show that $(2)$ does not apply.