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$f :\mathbb R \rightarrow \mathbb R$ 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?

Matthew Conroy
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Bag of Chips
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3 Answers3

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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.

shalop
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  • Why do we need $1 - \epsilon$ to be the bound, why isn't 1 sufficent? – Bag of Chips Feb 29 '16 at 19:16
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    Because if you look at the proof, the reason that $(x_k)$ is a Cauchy sequence follows from the fact that $\sum_1^{\infty} (1-\epsilon)^k$ is a summable sequence. This isn't true without the $\epsilon$. – shalop Feb 29 '16 at 19:18
  • $\color{red}{\text{"so it follows that $(x_k)$ is a Cauchy sequence"}}$ How can you said that $(1-\epsilon)^k|(x_1-x_0)|$ term can be arbitrary small$?$ I knew it's super late but can you help me to get that @Shalop Sir – emonHR Nov 02 '19 at 18:24
  • Since $1-\epsilon \in \left(0,1\right)$, we have that $(1-\epsilon)^k$ becomes arbitrarily small as $k \to \infty$. – Prasiortle May 13 '20 at 20:58
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Function $f(t)=t+\dfrac{1}{1+e^t}$ has no fixed point but $0<f'(t)<1$ for all real $t$.

RFZ
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The derivative bounded by 1 just gives the fact that if at all there exists a fixed point it is unique..to ensure that unique fixed point exists just make f a continuous function on any closed and bounded i.e compact interval. I suppose you dealing in the real line..

Upstart
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  • Uniqueness would follow by the mean value theorem. If $x_1<x_2$ with $f(x_1)=x_1$ and $f(x_2)=x_2$ then mean value theorem says that there is some $c\in (x_1,x_2)$ with $x_2-x_1=f(x_2)-f(x_1)=f'(c)(x_2-x_1)$, i.e, $f'(c)=1$. – shalop Feb 29 '16 at 19:14