$f:\mathbb R \to \mathbb R$ is a differentiable function such that $f'(x)\le r<1 $ , does $f$ necessarily have a fixed point ?

Question

Let $f:\mathbb R \to \mathbb R$ be a differentiable function . If $\exists r \in \mathbb R $ such that $|f'(x)|\le r<1 , \forall x \in \mathbb R$ then using Lagrange's theorem one can show $f$ is a Lipscitz contraction and then use Banach contraction principle to conclude $f$ has a unique fixed-point. My question is what happens if $f'(x)\le r<1 , \forall x \in \mathbb R$ ? Then does $f$ even have a fixed point ?

Answer 1

If $f$ has no fixed point then $f(x)<x$ for all $x$ or $f(x)>x$ for all $x$. Assume that $f(x)>x$ for all $x$, then $f(x)-f(0)>x-f(0)$ for all $x>0$. Take $x=f(0)t$ for $t>0$, then $f(f(0)t)-f(0)>f(0)(t-1)$. Dividing $f(0)t$ both sides then $$\frac{f(f(0)t)-f(0)}{f(0)t}\ge 1-\frac{1}{t}.$$ By mean value theorem then we can find for some $c\in (0,f(0)t)$ such that $f'(c)\ge 1-1/t$, so $f'$ cannot be bounded by $r<1$.

Similarly, you can check that $f'$ cannot be bounded by $r<1$ in the case $f(x)<x$ for all $x$.