I need to prove the following:
$f:[0,1]\to [0,1]$ continuous then exists $x_0\in [0,1]$ such that $f(x_0) = x_0$
$f$ continuous, then there are $x_1,x_2$ such that $f(x_1)=0$ and $f(x_2)=1$ and by the mean value theorem there exists $c\in [0,1]$ such that $0\le f(c)\le 1$ but I couldn't find any relation to what I want to prove. I also imagined that if $f(x)=x$ then it's proved, but if not, then there should exist one point $p$ such that $f'(p)=\frac{\sqrt{2}}{2}$ but that does not use mean value theorem. Any ideas on how to prove it?
Best Answer
First define $g(x)=f(x)-x$. Now $g(0)=f(0)-0=f(0)\geq 0$ and $g(1)=f(1)-1\leq 0$.
Now if either $g(0)=0$ or $g(1)=0$, then we have $f(0)=0$ or $f(1)=1$ respectively, so we can assume $g(0)>0$ and $g(1)<0$. Now $g$ is continuous on $[0,1]$ since $f$ is continuous, so we can apply the Intermediate Value Theorem, which gives us that $g(a)=0$ for some $a\in (0,1)$. Choosing $c=a$ we have that $$ f(c)=g(c)+c=c $$