Proving Bounded Function and Continuity: Intermediate Value Theorem Explained

[lepton123]

Homework Statement

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

Homework Equations

Intermediate value theorem?

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

 

[gopher_p]

Homework Statement

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

Homework Equations

Intermediate value theorem?

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

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 $g(x)=f(x)-x$.

 

[lepton123]

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

 

[gopher_p]

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

Yes.

 

[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)

 

[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.

 

[D H]

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)

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.