TITLE: Show $f$ has exactly one fixed point if $f'(x) <1$ for all $x$
QUESTION [5 upvotes]: I'm trying to use the mean value theorem to show that for differentiable $f:\mathbb{R} \to \mathbb{R}$, if $f'(x) < 1 \forall x\in\mathbb{R}$, then $f$ has exactly one fixed point. I know how to show that if $f'(x)\neq 1$ then $f$ can have at most one fixed point. I think I need to use that $f'(x) \leq 1$ to see that $f$ is decreasing, but I'm not sure how to use that.

REPLY [3 votes]: The thing you are trying to prove is false, even with strict inequality.
$$ f(x) = \frac{x + \sqrt {1+x^2}}{2} $$
has derivative below $1$ but no fixed point: we always have $f(x) > x$