$f :\mathbb R \rightarrow \mathbb R$ is differentiable on $\mathbb R $ and $|f'(x)| \lt 1$, does $f$ have a fixed point?
I think it does but I can't finish the proof.
Let's define $g(x) = f(x) – x$, we want to prove that this function is equal to $0$ at some point.
$g'(x) = f'(x) – 1 \in (-2,0)$, so $g$ is strictly decreasing. What now?
Best Answer
Maybe try $f(x) = \frac{1}{2}(x+\sqrt{x^2+1})$ for a counterexample.
However, the theorem becomes true if you replace $1$ by $1-\epsilon$, for some $\epsilon>0$. For that you can prove it like this: by the mean value theorem we see that $|f(x)-f(y)|<(1-\varepsilon)|x-y|$ for any $x,y \in \Bbb R$. So pick a point $x_0 \in \Bbb R$ and define $x_{k+1}=f(x_k)$. We see that $|x_{k+1}-x_k|=|f(x_k)-f(x_{k-1})| \leq (1-\epsilon)|x_k-x_{k-1}|$, so by induction $|x_{k+1}-x_k|<(1-\epsilon)^k|x_1-x_0|$, so it follows that $(x_k)$ is a Cauchy sequence, so it converges to some point $p\in \Bbb R$. Then by continuity of $f$, we see that $p=\lim x_k=\lim x_{k+1} = \lim f(x_k)=f(p)$.
BTW, this is known as the contraction mapping theorem.