Banach Fixed Point Theorem and the function $f(x)=x+e^{-x}$

Question

let $X = [0,\infty)$ equipped with the standard metric $d(x,y) = |x-y|$.

Let $f: X \rightarrow X$ be defined by $f(x)=x+e^{-x}$

Explain why this function doesn't contradict Banach's Fixed Point Theorem.

In the earlier parts of the problem I showed that:

$X$ is a complete metric space

$f$ is a contraction on $X$ (in the sense that $|f(x)-f(y)| \le |x-y|$ for all $x,y \in X$ )

$f$ doesn't have a fixed point in $X$.

My intuition is that the sort of contraction required for Banach's theorem is stronger than the contraction I've proven, but I can't seem to prove that the stronger contraction doesn't hold for $f$

Answer 1

The hypothesis of Picard's theorem which is not verified is that $f$ is NOT a contraction : you are a contraction when, for $k<1,$ $$\forall x,y\in\mathbb{R^+},|f(x)-f(y)|\leq k|x-y|.$$ Here it is not the case : you get indeed that $k=1$ is your optimal Lipschitz constant.