# F(x)=x for some x

1. Oct 14, 2013

### lepton123

1. The problem statement, all variables and given/known data

I have to prove that if a function is bounded for x between (0,1) and f(x) is bounded also between (0,1) and f(x) is continuos that there exists some x such that f(x)=x

2. Relevant equations

Intermediate value theorem?

3. The attempt at a solution

I know that this problem makes intuitive sense, and I've drawn out bounds, and I know obviously, that this statement will hold true, and the proof probably involves the intermediate theorem law, but I am just not sure how to construct my proof for it. Can I just say that for f(0)<x<f(1) then x=f(x) for some x (and repeat the same thing for my other two cases, when f(1)>f(0) and f(1)=f(0))?

Any help would be great

2. Oct 14, 2013

### gopher_p

First off, you need the conditions you've listed to hold on $[0,1]$. A counterexample on $(0,1)$ would be $f(x)=x^2$.

You are correct that the Intermediate Value Theorem might prove useful. Think about the function $g(x)=f(x)-x$.

3. Oct 14, 2013

### lepton123

Okay, I considered doing that; would I try to now show that g(x)=0 for some x?

4. Oct 14, 2013

### gopher_p

Yes.

5. Oct 14, 2013

### lepton123

Okay, I think I got it now, thanks!
I just considered two points; g(1) and g(0), and showed how between those points, there always exists a zero as g(1) can be between 0 and -1 and g(0) is between 0 and 1, so by the intermediate value theorem, there has to be a zero between these two points for any g(x) that satisfy the properties of f(x)

6. Oct 14, 2013

### gopher_p

That sounds like the right idea. You'll definitely need to beef up the rigor a bit to make it a proper proof.

7. Oct 14, 2013

### D H

Staff Emeritus
It's important to remember that that "between" is inclusive of rather than exclusive of the end points. You used (0,1) in the problem statement rather than [0,1]. gopher_p's counterexample of f(x)=x2 satisfies the conditions of the opening post, but there is no point in (0,1) where f(x)=x. There is such a point (two points in this case) if one looks at [0,1] instead.