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.

- April 11th 2011, 06:54 PMpaulrbQuestion 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. - April 11th 2011, 07:24 PMSambit
It is the well-known

**fixed point theorem**. Suppose is continuous on and for every .

If or , the theorem is proved. So we try to prove the theorem assuming

and .

Let for all . Hence and and is continuous on , that is, is an intermediate value of on . Hence by**intermediate value theorem**, there exists a point

such that --which means Hence the prrof.

EDIT: is equivalent to - April 11th 2011, 07:29 PMDrexel28