Universal chord theorem

A chord (in red) of length 0.3 on a sinusoidal function. The universal chord theorem guarantees the existence of chords of length 1/n for functions satisfying certain conditions.

In mathematical analysis, the universal chord theorem states that if a function f is continuous on [a,b] and satisfies $f(a)=f(b)$, then for every natural number $n$, there exists some $x\in [a,b]$ such that $f(x)=f(x+\frac{b-a}{n})$.

History

The theorem was published by Paul Lévy in 1934 as a generalization of Rolle's theorem.

Statement of the theorem

Let $H(f)=\{h\in [0,+\mathrm{\infty }):f(x)=f(x+h)\text{ for some }x\}$ denote the chord set of the function f. If f is a continuous function and $h\in H(f)$, then $\frac{h}{n}\in H(f)$ for all natural numbers n.

Case of n = 2

The case when n = 2 can be considered an application of the Borsuk–Ulam theorem to the real line. It says that if $f(x)$ is continuous on some interval $I=[a,b]$ with the condition that $f(a)=f(b)$, then there exists some $x\in [a,b]$ such that $f(x)=f(x+\frac{b-a}{2})$.

In less generality, if $f:[0,1]\to ℝ$ is continuous and $f(0)=f(1)$, then there exists $x\in [0,\frac{1}{2}]$ that satisfies $f(x)=f(x+1/2)$.

Proof of n = 2

Consider the function $g:[a,\frac{b+a}{2}]\to ℝ$ defined by $g(x)=f(x+\frac{b-a}{2})-f(x)$. Being the sum of two continuous functions, $g$ is continuous, $g(a)+g\left(\frac{b+a}{2}\right)=f(b)-f(a)=0$. It follows that $g(a)\cdot g\left(\frac{b+a}{2}\right)\le 0$ and by applying the intermediate value theorem, there exists $c\in [a,\frac{b+a}{2}]$ such that $g(c)=0$, so that $f(c)=f(c+\frac{b-a}{2})$. This concludes the proof of the theorem for $n=2$.

Proof of general case

The proof of the theorem in the general case is very similar to the proof for $n=2$ Let $n$ be a non negative integer, and consider the function $g:[a,b-\frac{b-a}{n}]\to ℝ$ defined by $g(x)=f(x+\frac{b-a}{n})-f(x)$. Being the sum of two continuous functions, $g$ is continuous. Furthermore, $\sum _{k=0}^{n-1}g(a+k\cdot \frac{b-a}{n})=0$. It follows that there exists integers $i,j$ such that $g(a+i\cdot \frac{b-a}{n})\le 0\le g(a+j\cdot \frac{b-a}{n})$ The intermediate value theorems gives us c such that $g(c)=0$ and the theorem follows.

Counterexample for non-integer n

Let $r\in ℝ$ be arbitrary, and consider the function $f:[0,1]\to ℝ$ defined by $f(x)={\sin }^2\left(\frac{\pi x}{r}\right)-x{\sin }^2\left(\frac{\pi }{r}\right)$. It is immediate that $f$ is continuous, and $f(0)=f(1)=0$. If some $x\in [0,1]$ satisfies $f(x)=f(x+r)$, then $r{\sin }^2\left(\frac{\pi }{r}\right)=0$ which implies that $r=\frac{1}{n}$ for some integer $n$. Therefore the theorem does not hold for non-integer values of $n$. Template:Citation needed