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$f$ 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.

  • 3
    possible duplicate of [Horizontal chord of length $\frac{1}{2}$ in the graph of a continuous function.](http://math.stackexchange.com/questions/16374/horizontal-chord-of-length-frac12-in-the-graph-of-a-continuous-function)2011-03-01

3 Answers 3

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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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    $f$ is assumed only to be continuous.2011-02-28
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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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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 ]