# Thread: Question involving a continuous function on a closed interval

1. ## Question involving a continuous function on a closed interval

Intuitively this is obvious by graphing g(x) = x on [0,1] and seeing since f is continuous it has to intersect with g at some point. But I spent a long time and cannot figure out how to prove this.

2. It is the well-known fixed point theorem. Suppose $f$ is continuous on $[0,1]$ and $f(x)\epsilon[0,1]$ for every $x\epsilon[0,1]$.

If $f(0)=0$ or $f(1)=1$, the theorem is proved. So we try to prove the theorem assuming

$f(0)>0$ and $f(1)<1$.

Let $g(x)=f(x)-x$ for all $x\epsilon[0,1]$. Hence $g(0)>0$ and $g(1)<0$ and $g$ is continuous on $[0,1]$, that is, $0$ is an intermediate value of $g$ on $[0,1]$. Hence by intermediate value theorem, there exists a point

$c\epsilon(0,1)$ such that $g(c)=0$ --which means $f(c)=c.$ Hence the prrof.

EDIT: $c$ is equivalent to $x_0$

3. Originally Posted by paulrb

Intuitively this is obvious by graphing g(x) = x on [0,1] and seeing since f is continuous it has to intersect with g at some point. But I spent a long time and cannot figure out how to prove this.
Alternatively, suppose that $f(x)\ne x$ then the mapping $\displaystyle f:[0,1]\to\{-1,1}:\frac{|f(x)-x|}{f(x)-x}$ is a continuous surjection which is impossible.