# Universal Chord Theorem

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.

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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The statement should seem intuitive to you, and when it does the proof should follow fairly easily. What we have here (can be thought of as) a loop in $\mathbb{R}^2$ parameterized by $t \in [0,1]$. Picture the values of $f(a)$ and $f(a+ 1/2)$ as $a$ varies continuously from 0 to 1/2. Can you visualize why these values must equal each other at some point? Hope this at least makes sense and possibly helps :) – jericson Jan 4 '11 at 23:23
The function f(x) = x^2 - x satisfies the problem assumptions but isn't periodic. – Qiaochu Yuan Jan 5 '11 at 0:47

You want to use the intermediate value theorem, but not applied to $f$ directly. Rather, let $g(x)=f(x)-f(x+1/2)$ for $x\in[0,1/2]$. You want to show that $g(a)=0$ for some $a$. But $g(0)=f(0)-f(1/2)=f(1)-f(1/2)=-(f(1/2)-f(1))=-g(1/2)$. This gives us the result: $g$ is continuous and changes sign, so it must have a zero.

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You define $g(x)=f(x) - f(x+1/2)$ for $x \in [0,1/2]$ and you get that $g(0)=-g(1)$. Why $a \in [0,1/2]$ then? – Ma.H Jan 4 '11 at 23:26
@Andres: I vaguely recall seeing an extension of this to show what values of a (besides 1/2) you could prove that you could find an x such that f(x)=f(x+a). I think this goes easily to 1/n for any n, but there may be more. – Ross Millikan Jan 4 '11 at 23:28
@Ma.H: It's a typo, should be $g(0)=-g(1/2)$. – Shai Covo Jan 4 '11 at 23:35
(Sorry about the typo.) @Ross: I don't know, it doesn't ring a bell but sounds interesting. I'll see if I can think of a non-trivial extension. – Andrés E. Caicedo Jan 5 '11 at 0:39
@Ross: $1/n$ is right, and those are the only ones. See my answer. – Aryabhata Jan 5 '11 at 6:21

Interestingly,

The numbers of the form $r = \displaystyle \frac{1}{n} \ \ n \ge 1$ are the only numbers such that for any continuous function $\displaystyle f:[0,1] \to \mathbb{R}$ such that $\displaystyle f(0) = f(1)$, there is some point $\displaystyle c \in [0,1]$ such that $\displaystyle f(c) = f(c+r)$.

For any other $r$ we can find a continuous function for which there is no $\displaystyle c$ such that $\displaystyle f(c) = f(c+r)$.

For a proof that $\displaystyle r = \frac{1}{n}$ satisifies this property, let $\displaystyle g(x) = f(x) - f(x+ \frac{1}{n})$

Then we have that $\displaystyle \sum_{k=0}^{n-1} \ g\left(\frac{k}{n}\right) = 0$.

Thus if none of $\displaystyle g\left(\frac{k}{n}\right)$ are $\displaystyle 0$, then there are some $\displaystyle i,j$ such that $\displaystyle g\left(\frac{i}{n}\right) \gt 0$ and $\displaystyle g\left(\frac{j}{n}\right) \lt 0$.

For any other $\displaystyle r \in (0,1)$

Consider the following example, due to Paul Levy.

$\displaystyle f(x) = \sin^2\left(\frac{\pi x}{r}\right) - x \ \sin^2\left(\frac{\pi}{r}\right)$

If we have $\displaystyle f(x) = f(x+r)$, then we have that $\displaystyle r\ \sin^2\left(\frac{\pi}{r}\right) = 0$ and hence $\displaystyle r = \frac{1}{m}$ for some integer $\displaystyle m$.

Apparently this is called the Universal Chord Theorem (due to Paul Levy!).

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I like this example! – Andrés E. Caicedo Jan 5 '11 at 6:21
Awesome, yaar. – user9413 May 25 '11 at 4:28
@Chandru: Thanks, but credit goes to Paul Levy :-) – Aryabhata May 25 '11 at 4:29
Credit goes to you for knowing his theorem as well :) – user9413 May 25 '11 at 4:36

Hint: consider $g(x)$, defined on $[0,1/2]$ by $g(x)=f(x+1/2)-f(x)$

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## protected by AryabhataFeb 26 '12 at 10:41

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