# Real Analysis, continuity on [0,1]

• Oct 7th 2008, 02:18 PM
pila0688
Real Analysis, continuity on [0,1]
Q: let f be continuous from [0,1] to [0,1]. Show their is an x on [0,1] for which f(x) = x

Our hint was to use g(x) = f(x) - x and use the IVT

We figure that since our Range is from [0,1] with our max = 1 and min = 0 that if we could show that the function attains a value above y=x and one below y=x, that continuity implies that it mus cross y=x attaining our value.

Problem is how do we do this, my intial thinking was using the defintition of continuity using our

|f(x) - f(a)| < epsilon where we take 2 cases where f(a) is our min and f(a) is our max, where working it down would take us to

|f(x) - f(max)| < epsilon
|f(x) - f(min)| < epsilon

|f(x) - 1| < epsilon
|f(x) - 0| < epsilon

putting our x

|f(x)| < x < |f(x) - 1|

hence where I stopped and got stuck. Thanks for any help
• Oct 7th 2008, 02:58 PM
ThePerfectHacker
Quote:

Originally Posted by pila0688
Q: let f be continuous from [0,1] to [0,1]. Show their is an x on [0,1] for which f(x) = x

Our hint was to use g(x) = f(x) - x and use the IVT

Note $\displaystyle g(0) = f(0) - 0 = f(0)\geq 0$ and $\displaystyle g(1) = f(1) - 1 \leq 1 -1 = 0$. Therefore there is $\displaystyle 0\leq x_0\leq 1$ so that $\displaystyle g(x_0) = 0 \implies f(x_0) = x_0$.

This is Mine 1:)8:):)th Post!!!
• Oct 7th 2008, 03:01 PM
icemanfan
(1) Suppose f(a) - a < 0 but f(b) - b > 0 for some 0 < a < 1, 0 < b < 1.

Then by the IVT, there is c such that a < c < b, and hence 0 < c < 1, with f(c) - c = 0.

(2) Suppose f(x) - x > 0 for all 0 <= x <= 1. Then f(1) - 1 > 0, or f(1) > 1, which is impossible.

(3) Suppose f(x) - x < 0 for all 0 <= x <= 1. Then f(0) - 0 < 0, or f(0) < 0, which is impossible.

Then if situation (1) does not exist, either f(x) - x <= 0 for all 0 <= x <= 1 or
f(x) - x >= 0 for all 0 <= x <= 1, with equality occurring at least once.
• Oct 9th 2008, 11:40 AM
bkarpuz
Quote:

Originally Posted by pila0688
Q: let f be continuous from [0,1] to [0,1]. Show their is an x on [0,1] for which f(x) = x

Our hint was to use g(x) = f(x) - x and use the IVT

We figure that since our Range is from [0,1] with our max = 1 and min = 0 that if we could show that the function attains a value above y=x and one below y=x, that continuity implies that it mus cross y=x attaining our value.

Problem is how do we do this, my intial thinking was using the defintition of continuity using our

|f(x) - f(a)| < epsilon where we take 2 cases where f(a) is our min and f(a) is our max, where working it down would take us to

|f(x) - f(max)| < epsilon
|f(x) - f(min)| < epsilon

|f(x) - 1| < epsilon
|f(x) - 0| < epsilon

putting our x

|f(x)| < x < |f(x) - 1|

hence where I stopped and got stuck. Thanks for any help

This is called as Fixed Point Theorem.