Universal Chord Theorem
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Let $f in C[0,1]$ and $f(0)=f(1)$. How do we prove $exists a in [0,1/2]$ such that $f(a)=f(a+1/2)$? In fact, for every positive integer $n$, there is some $a$, such that $f(a) = f(a+frac{1}{n})$. For any other non-zero real $r$ (i.e not of the form $frac{1}{n}$), there is a continuous function $f in C[0,1]$, such that $f(0) = f(1)$ and $f(a) neq f(a+r)$ for any $a$. This is called the Universal Chord Theorem and is due to Paul Levy. Note: the accepted answer answers only the first question, so please read the other answers too, and also this answer by Arturo to a different question: https://math.stackexchange.com/a/113471/1102 This is being repurposed in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions. and here: List of abstract duplicates.
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