Let $ f:I \rightarrow I$, with $I$closed subset of the real line, be a contraction, meaning that there exists a constant $L \in (0,1)$ such that $\forall x, \forall y \in I, |f(x)-f(y)|\leq L|x-y| $. Then there exists a unique point $z \in I$ such that $f(z)=z$. Now, i get that if we only require that $|f(x)-f(y)|<|x-y| $ instead, the theorem does not work (take for example $x+\frac{1}{x} $ in $ [1, +\infty]$).
But what if the domain is compact? I can't come up with a counterexample to the theorem.
Thanks in advance.
