LEt $X\subseteq \mathbb{R}^d$ be a closed set, and $f: X \to X$ be a function such that
$$ ||f(x_1) – f(x_2)|| < ||x_1 – x_2|| \; \; \forall x_1, x_2 \in X $$
If $X$ is compact, then there exists a unique $p \in X$ such that $f(p) = p $.
How Can I approach this problem? My thought was to assume by contradiction that $\forall x $, $f(x) \neq x$. But, then how can I obtain a contradiction?
Best Answer
The function $x\mapsto \lVert x-f(x)\rVert$ is continuous. Since $X$ is compact, it attains its minimum, say in $x_0\in X$. What follows for $x_0$ and $f(x_0)$?