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Suppose $f$ is continuous on $[0,2]$ and that $f(0) = f(2)$. Then $\exists$ $x,$ $y \in [0,2]$ such that $|y - x| = 1$ and that $f(x) = f(y)$.

APPROACH: Let $g(x) = f(x+1) - f(x)$ on $[0,1]$. Then $g$ is continuous on $[0,1]$, and hence $g$ enjoys the intermediate value property!. Now notice $$g(0) = f(1) - f(0)$$ $$g(1) = f(2) - f(1)$$ Therefore $$ g(0)g(1) = -(f(0) - f(1))^2 < 0$$ since $f(0) = f(2)$. Therefore, there exists a point $x$ in $[0,1]$ such that $g(x) = 0$ by the intermediate value theorem. Now, if we pick $y = x + 1$, i think the problem is solved.

I would like to ask you guys for feedbacks. Is this solution correct? Is there a better way to solve this problem? Thanks!

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You might have $g(0)g(1)=0$ and not $<0$, but then $x=0$, $y=1$ is a solution. – Hagen von Eitzen Oct 22 '12 at 21:29
Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Belgi Oct 22 '12 at 21:29
up vote 2 down vote accepted

Very good!

It is enough to write $g(1)=-g(0)$..

And, as Hagen commented, you only forgot to mention the case when already $f(0)=f(1)$, but then it is readily done.

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