Universal chord theorem

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 $f(x)=f\left(x+{\frac {b-a}{n}}\right)$ .^[1]

The theorem was published by Paul Lévy in 1934 as a generalization of Rolle's theorem.^[2]

Statement of the theorem

Let $H(f)=\{h\in [0,+\infty ):f(x)=f(x+h){\text{ for some }}x\}$ 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. ^[3]

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 $f(x)=f\left(x+{\frac {b-a}{2}}\right)$ .

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

Proof of n = 2

Consider the function $g:\left[a,{\dfrac {b+a}{2}}\right]\to \mathbb {R}$ defined by $g(x)=f\left(x+{\dfrac {b-a}{2}}\right)-f(x)$ . Being the sum of two continuous functions, $g$ is continuous, $g(a)+g\left({\dfrac {b+a}{2}}\right)=f(b)-f(a)=0$ . It follows that $g(a)\cdot g\left({\dfrac {b+a}{2}}\right)\leq 0$ and by applying the intermediate value theorem, there exists $c\in \left[a,{\dfrac {b+a}{2}}\right]$ such that $g(c)=0$ , so that $f(c)=f\left(c+{\dfrac {b-a}{2}}\right)$ . 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 $g:\left[a,b-{\dfrac {b-a}{n}}\right]\to \mathbb {R}$ defined by $g(x)=f\left(x+{\dfrac {b-a}{n}}\right)-f(x)$ . Being the sum of two continuous functions, $g$ is continuous. Furthermore, $\sum _{k=0}^{n-1}g\left(a+k\cdot {\dfrac {b-a}{n}}\right)=0$ . It follows that there exists integers $i,j$ such that $g\left(a+i\cdot {\dfrac {b-a}{n}}\right)\leq 0\leq g\left(a+j\cdot {\dfrac {b-a}{n}}\right)$ The intermediate value theorems gives us c such that $g(c)=0$ and the theorem follows.

Counterexample for non-integer n

Let $r\in \mathbb {R}$ be arbitrary, and consider the function $f:[0,1]\to \mathbb {R}$ 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 $n$ . ^{[citation needed]}