Apostol Analysis Problem 6.5 Prove that $f(1)>1$ Let $f$ be a real function defined over $[0,1]$ such that $f(0)>0$, $f(x) \neq x$ for all $x$ and if $x \leq y$ then $f(x) \leq f(y)$. Let $A=\{x \ : \ f(x)>x\}$ Prove that $\sup A \in A$ and that $f(1)>1$
I have already proved that $\sup A \in A$ but I don't know how to prove that $f(1)>1$. I was trying to prove that $1= \sup A$. Since $a \in (0,1]$ then $a \leq 1$, so all I need to do is to prove that $1 \leq a$. Could someone help me with a hint?
 A: If $a=\sup A$ but $f(a)\le 1$, then $f(a)>a$ implies $f(f(a))>f(a)$, hence $f(a)\in A$. For any $x\in A$, we have $x\le a<f(a)$ so $f(a)$ is both contained in $A$ and is a strict upper bound of $A$, which is a contradiction.
Therefore, $f(a)>1$ is forced.
$f(a)>a$ implies $f(f(a))\ge f(a)$ since $f$ is weakly increasing, but $f(f(a))\neq f(a)$ holds because $f(x)\neq x$ always. Then $f(f(a))>f(a)$ is true.
A: We claim that $a:=\sup A=1. $ Assume by contradiction that $a<1.$  We have $f(a)>a. $ Therefore $f(a)>1,$ as otherwise $f(a)\in A.$ Hence $f(1)\ge f(a)>1.$
The argument $1$ may be replaced by $0<u<1.$ It can be done by adjusting the proof or by change of variables as follows.
For a fixed $0<u<1,$ let $g(x)={1\over u}f(ux).$ Then $g(0)>0$ and  $g(x)\neq x$ iff $f(x)\neq x.$ By the first part we get $g(1)>1,$ hence $f(u)>u.$
A: Claim: We have $A=[0,1]$.
Proof: Assume that $[0,1]\setminus A\neq\emptyset$ and let $x^*\in[0,1]\setminus A=\{x\in[0,1]:f(x)\le x\}$. Since we have $f(x)\neq x$, we have $f(x^*)<x^*$. Now, let $x^\circ=\inf\{x\in[0,1]:f(x)<x\}$. Notice that $x^\circ\ge f(0)>0$ since $f(x)\ge f(0)$. Assume that $f(x^\circ)>x^\circ$, take a sequence $x_n\searrow x^\circ$ with $f(x_n)<x_n$ and notice that we hence eventually have $f(x_n)<f(x^\circ)$, which is a contradiction to $f$ being non-decreasing. So, we have $f(x^\circ)<x^\circ$. But for all $x<x^\circ$ we have $f(x)>x$, so for $x<x^\circ$ sufficiently close to $x^\circ$ we have $f(x)>f(x^\circ)$, which is a contradiction again. This shows that $A=[0,1]$.
In particular, we have $\sup A\in A$ and $f(1)>1$.
