Prove that if $F:[a,b]\to \mathbb{R}$, if $F'$ is continuous, and if $|F'(x)|<1$ on $[a,b]$, then $F$ is a contraction. Does $F$ necessarily have

Question

Prove that if $F:[a,b]\to \mathbb{R}$, if $F'$ is continuous, and if $|F'(x)|<1$ on $[a,b]$, then $F$ is a contraction. Does $F$ necessarily have a fixed point?

Let $x,y\in [a,b]$, by the mean value theorem, there exists a $c\in[x,y]$ such that $F(x)-F(y)=F'(c)(x-y)$, with which $|F(x)-F(y)|=|F'(c)||x-y|$ and so if $\lambda:=|F'(c)|$ then $\lambda<1$, then $|F(x)-F(y)|=|F'(c)||x-y|\leq \lambda |x-y|$ and so $F$ is contractive.

For the second question I think that $F$ does have a fixed point since $[a,b]$ is closed and $\mathbb{R}$ is closed, in addition $F([a,b])\subset \mathbb{R}$.

Is this reasoning correct? Thank you very much.

Answer 1

Your reasoning is not quite right for the first question. You nowhere used continuity of $F'$ in your answer. Kind of suspicious, right? To fix this, note $|F'|$ is continuous on $[a,b].$ It follows that $|F'|$ attains a maximum value somewhere in $[a,b],$ say at $x_0.$ We are given that $|F'|$ is everywhere less than $1,$ so we have $|F'(x_0)|<1.$ It follows that for any $x,y\in [a,b],$

$$|F(y)-F(x)| = |F'(c(x,y))(y-x)| \le |F'(x_0)||y-x|.$$

Since $|F'(x_0)|<1$ and the above works for all $x,y\in [a,b],$ $F$ is a contraction.

Answer 2

Consider $F:[0,1]\rightarrow\mathbb{R}$ defined by $F(x)=x/2+10$ its image is $[10,10+1/2]$, $F$ is contraction and does not have a fixed point.

Answer 3

$$f(x)=\frac {x-1}{2}$$ on [1,2] satisfies your conditions but does not have a fixed point.