# Continuity problem

• Sep 28th 2013, 09:13 AM
Orpheus
Continuity problem
Hello MHF, please help on this problem; (and by the way, we only did the continuity lesson )
We are given :
$f:[0,1]\rightarrow [0,1]$
$g:[0,1]\rightarrow [0,1]$
both $f$ and $g$ are continuious on $[0,1]$
$f(g(x)) = g(f(x))$
If $f(a) = a$ then $f(g(a)) = g(a)$

Show that $\exists \alpha \in [0,1] f(\alpha) = g(\alpha)$

P.S : use proof by contradiction
• Sep 28th 2013, 09:35 AM
Orpheus
Re: Continuity problem
I tried to do this
let h(x) = f(x) - g(x)
for every x in [0,1] h(x)≠0
then for every x in [0,1] h(x) > 0 ( or < 0)
but after this i could't find a contradiction
• Sep 28th 2013, 11:52 AM
Orpheus
Re: Continuity problem
Please I need to solve it before Monday morning.
• Sep 28th 2013, 05:38 PM
johng
Re: Continuity problem
Hi,
Since you need this by Monday morning, I can only assume it's either an exam question or homework for credit. So here are some hints:
Let a be a fixed point of f (f(a)=a) -- you need to prove why a exists. Now consider the sequence x1 = a and xn+1 = g(xn).

By the way, the above leads to a direct proof, not a proof by contradiction.
• Sep 28th 2013, 09:01 PM
chen09
Re: Continuity problem
To start you off, suppose without loss of generality
$f(x)>g(x) \quad \text{ for an open interval in } [0,1]$
You'll get a contradiction along the way.

Note: Using the auxiliary function h you defined, could be used for a direct proof, but at that point you guys are assumed to know Intermediate Value Theorem, etc.

Disclaimer: If my line of thought is wrong or it has missing holes, please correct and/or inform me.
• Oct 1st 2013, 05:35 AM
johng
Re: Continuity problem
Hi again,
If you're still interested after your deadline, here's a complete solution:

Attachment 29357

Chen, I'm interested in seeing a proof along the lines suggested by your post. Could you provide such?