Question:
Let $f:[0;1] \rightarrow [0;1]$ a strictly increasing fct. Prove that there exists at least one $x \in [0;1]$ s.t. $f(x)=x$.
The prove is false read selected answer for further explanation
My answer:
First by absurd let's suppose that it exist a function $f(x) \neq x \; \forall x \in [0;1]$ which is strictly increasing and define as given in the question.
That means especially that $ 0 \leq f(x) \leq 1 $.
(i) – First let prove, wlog, that such function $f(x)$ can not satisfy $f(x)<x \; \forall x \in [0;1]$. Indeed in such a case in the point $x=0$ the function has to satisfy $f(0)<0$ and that impossible because the question explicitly states that: $f(x) \geq 0 \; \forall \; x \in [0;1]$
The proof is very similar to prove that such function $f(x)$ can not satisfy $f(x)>x \; \forall x \in [0;1]$.
(ii) – Now let write the two set $A_>=\left \{ x \in [0;1] \; | f(x)>x \right \}$ and $A_<=\left \{ x \in [0;1] \; | f(x)<x \right \}$. Obviously $x=0 \in A_>$ and $x=1 \in A_<$.
Because in the other case we will get wlog: $f(1)>1$ while $f(x) \leq 1$ by assumption.
(iii) – Now let choose $x_1 \in A_>$ and $x_2 \in A_<$ s.t. $x_1<x_2$ (exists according to (ii) ) and let define the two following sequence $a_n$ and $b_n$ as follow:
- $a_1=x_1$ , $a_2=f(x_1)=f(a_1)$ , … , $a_{n} = f(a_{n-1}) \; \Rightarrow a_{n-1}< a_n = f(a_{n-1})$
We should remark that $a_n \in A_>$ too as it is defined from $a_{n-1} \in A_>$
Indeed i.e. $f(a_2)=f(f(x_1)) > f(x_1) = a_2$ as $f(x_1)>x_1$ and the fct is strictly increasing, hence $\forall \; b>a \Rightarrow f(b)>f(a)$ so in particular if $b=f(x_1)>a=x_1$ (which is true as $x_1 \in A_>$. - $b_1=x_2$ , $b_2=f(x_2)=f(b_1)$ , … , $b_{n} = f(b_{n-1}) \; \Rightarrow b_{n-1}< b_n = f(b_{n-1})$
Similar remark for $b_n$ to those for $a_n$.
Conclusion:
Hence: $a_{n} = f(a_{n-1})$ and $b_{n} = f(b_{n-1}) \Rightarrow a_{n-1}< a_n = f(a_{n-1}) < b_{n-1}< b_n = f(b_{n-1})$ That means that $a_n$ and $b_n$ are two strictly increasing or decreasing bounded sequences. So they have a limit.
So concerning $a_n$ we get on one side: $\lim_{n \to \infty }a_n = l \in (0;1) $ and on the other side: $\lim_{n \to \infty }a_n = \lim_{n \to \infty }f(a_{n-1}) = f(l)$ so by the uniqueness of the limit we get: $f(l)=l$ and that there exists a point $l \in [0;1]$ that is the limit of $a_n$ and $f(a_{n-1})$ and that satisfies (by the uniqueness of the limit of a sequence) $f(l)=l$
Q.E.D.
Is it correct? If anyone has a better solution feel free to post it. Thank for your help.
Best Answer
No, your proof is incorrect.
A counterexample
Here is a summary of your proof. Suppose $f(a_1)>a_1$. Let $a_n=f(a_{n-1})$. Then $a_n$ converges to some number $l$ when $n\to\infty$. Then $f(l)=l$.
Here is a counterexample to your proof.
Consider function $f(x)=\begin{cases}\frac14+\frac x2&\text{if }x\in[0;\frac12)\\\frac34+\frac{x-\frac12}2&\text{if }x\in[\frac12;1]\end{cases}$
Since $f(0)=\frac14>0$, let $a_1=0$ in your proof. The sequence of $a_i$'s is $0,\frac14, \frac38, \frac7{16}, \frac{15}{32},\cdots$. Check that $a_n=\frac12-\frac{1}{2^n}$ for all $n$. $l=\lim_{n\to\infty}a_n=\frac12$.
All are fine so far. However, $f(l)=\frac34\not=l$.
The mistake in the proof is, as pointed by FShrike, the following equality does not hold necessarily,
$$ \lim_{n \to \infty }f(a_{n-1}) = f(l)$$ since $f$ may not be continuous at $l$.
A correct proof
I will assume $f$ is weakly increasing only. A strictly increasing function is also weakly increasing.
Let $A_\ge=\{x\in[0;1]\mid f(x)\ge x\}\ni 0$. Since $A_\ge$ is a non-empty set bounded from above, its least upper bound is well-defined. . Let it be $\ell$.
For any $a\in A_\ge$, $\ell\ge a$. So $f(\ell)\ge f(a)\ge a$, i.e., $f(\ell)$ is also an upper bound for $A_\ge$. Hence $f(\ell)\ge\ell$.
Applying $f$ to both sides, we get $f(f(\ell))\ge f(\ell)$, i.e., $f(\ell)\in A_\ge$. Since $\ell$ is an upper bound for $A_\ge$, $\ell\ge f(\ell)$.
So $f(\ell)=\ell$.