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A function $f(x)$ is defined and continuous on the interval $[0,2]$ and $f(0)=f(2)$. Prove that the numbers $x,y$ on $[0,2]$ exist such that $y-x=1$ and $f(x) = f(y)$.

I can already guess this is going to involve the intermediate value theorem. So far I've defined things as such: I'm looking to satisfy the following conditions for values x, y:

  1. $f(x) = f(x+1)$
  2. $f(x) = f(y)$

I've defined another function, $g(x)$ such that $g(x) = f(x+1) - f(x)$ If I can show that there exists an $x$ such that $g(x) = 0$ then I've also proven that $f(x) = f(x+1)$.

since I'm given the interval [0,2], I can show that:

$g(1) = f(2) - f(1)$,

$g(0) = f(1) - f(0)$

I'm told that $f(2) = f(0)$ so I can rearrange things to show that $g(1) = f(0) - f(1) = -g(0)$. Ok, So i've shown that $g(0) = -g(1)$

How do I tie this up? I'm not able to close this proof. I know I need to incorporate the intermediate value theorem which states that if there's a point c in $(a,b)$ then there must be a value $a<k<b$ such that $f(k) = c $ because there's nothing else.

I thought maybe to use Rolle's theorem to state that since $f(0) = f(2)$ I know this function isn't monotonic. And if it's not monotonic it must have a "turning point" where $f'(x) = 0$ but it's not working out. Anyway I need help with this proof in particular and perhaps some advice on solving proofs in general since this type of thing takes me hours.


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+1, despite the lame title :) – David Mitra Feb 1 '12 at 18:53
up vote 5 down vote accepted

if $g(0)$ is positive, $g(1)$ will be negative and vice versa, so the IVT provides a root. if both are zero, $g(0)=g(1)=0=f(1)-f(0)=f(2)-f(1)$ and you're done as well.

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Hint: You are told that $f(2)=f(0)$ (not $f(1)$ as you say). This gives you that $g(1)=-g(0)$

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