# Question involving a continuous function on a closed interval

• Apr 11th 2011, 06:54 PM
paulrb
Question involving a continuous function on a closed interval
http://i.imgur.com/VgMah.png

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.
• Apr 11th 2011, 07:24 PM
Sambit
It is the well-known fixed point theorem. Suppose $\displaystyle f$ is continuous on $\displaystyle [0,1]$ and $\displaystyle f(x)\epsilon[0,1]$ for every $\displaystyle x\epsilon[0,1]$.

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

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

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

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

EDIT: $\displaystyle c$ is equivalent to $\displaystyle x_0$
• Apr 11th 2011, 07:29 PM
Drexel28
Quote:

Originally Posted by paulrb
http://i.imgur.com/VgMah.png

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 $\displaystyle f(x)\ne x$ then the mapping $\displaystyle \displaystyle f:[0,1]\to\{-1,1}:\frac{|f(x)-x|}{f(x)-x}$ is a continuous surjection which is impossible.