# For a continuous function, $f(0) = f(1) \Longrightarrow f(a) = f(a + 1/2)$

$f$ is continuous between $[0,1]$, and $f(0)=f(1)$.

I want to prove that there is an $a \in [0,0.5]$ such that $f(a+0.5)=f(a)$.

ok, so Rolle's theorem can be useful here, but I can't see the connection to the derivative,

(Weierstrass, Uniform continuity?) I'll be glad to instructions.

Thanks.

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Hint: the function is not injective, then, as you say, use Rolle's theorem. – Andy Feb 28 '11 at 14:11
You don't know if the function has a derivative, so in particular you can't assume anything on the derivative. – Yuval Filmus Feb 28 '11 at 14:13
@Yuval, right...I would have argumented this only using the fact that it is not injective. Thanks for the correction ;) – Andy Feb 28 '11 at 14:15

Consider the function $g(a) = f(a+0.5) - f(a)$ on the interval $[0,0.5]$, and use the intermediate value theorem.

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Thank you yuval. – user6163 Feb 28 '11 at 14:18
$f$ is assumed only to be continuous. – lhf Feb 28 '11 at 14:19

HINT: Work with the function $g(a)=f(a+0.5)-f(a)$. Consider $g(0)$ and $g(0.5)$ (and their sum).

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Thank you shai. – user6163 Feb 28 '11 at 14:17

Or

consider if there is a $b\in [0,1]$ such that $f(b) = f(0) = f(1)$.

What if there is no such $b$?

What if $b = 0.5$?

[ You don't need Rolle's theorem this way ]

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