# If $f:[a,b]\to[a,b]$ is a nondecreasing function then $\exists x_0\in[a,b]$ such that $f(x_0)=x_0$

I got this problem:

Prove that if $f:[a,b]\to[a,b]$ is a nondecreasing function then $\exists x_0\in[a,b]$ such that $f(x_0)=x_0$ (i.e. $f$ has a fixed point).

(Hint: set $A=\{x\in[a,b]|x\leq f(x)\}$ and show that $x_0=\sup f([a,b])$ exist and that $f(x_0)=x_0$)

I tried to show that $A\neq\emptyset$ by supposing that $A=\emptyset$ and trying to reach a contradiction, but I got stuck.

Thanks only help.

• Note that $a \in A$, so $A \neq \emptyset$. – Crostul Oct 4 '14 at 8:24
• Thanks, I totally missed it. – MathNerd Oct 4 '14 at 8:30
• You've got your hint wrong. It should read $x_0 = \sup f(A)$. – posilon Oct 4 '14 at 9:51

$A\not =\emptyset.$ For example, $a\in A$ since $f(a)\in [a,b]$, then $a \le f(a)$.