# 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.

Interestingly,

The numbers of the form $r = \displaystyle \frac{1}{n} \ \ n \ge 1$ are the only positve 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 $\displaystyle c \in [0,1-r]$ such that $\displaystyle f(c) = f(c+r)$.

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

For a proof that $\displaystyle r = \frac{1}{n}$ satisifies this property, let $\displaystyle g(x) = f(x) – f(x+ \frac{1}{n})$, for $\displaystyle x \in [0, 1-\frac{1}{n}].$

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 $\displaystyle \exists i,j \in [0, 1, …, n -1] \ni \displaystyle g\left(\frac{i}{n}\right) \gt 0$ and $\displaystyle g\left(\frac{j}{n}\right) \lt 0$.

For any positive $\displaystyle r$, consider the following example, due to Paul Levy.

$\displaystyle f(x) = \sin^2\left(\frac{\pi x}{r}\right) – x \ \sin^2\left(\frac{\pi}{r}\right)$. Clearly, $f$ is continuous and $f(0)=0=f(1).$

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

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