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

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

Statement of the theorem
Let $$ H(f) = \{ h \in [0, +\infty) : f(x) = f(x+h) \text{ for some } x \} $$ 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 $$ f(x) = f\left(x + \frac{b-a}{2}\right) $$.

In less generality, if $$ f : [0,1] \rightarrow \R $$ 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, $$g(a) + g(\dfrac{b+a}{2}) = f(b) - f(a) = 0$$. It follows that $$ g(a)\cdot g(\dfrac{b+a}{2})\le 0$$ 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 $$ f(c) = f\left(c + \dfrac{b-a}{2}\right)$$. Which 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 $$ g\left(a+i\cdot\dfrac{b-a}{n}\right)\le 0\le g\left(a+j\cdot\dfrac{b-a}{n}\right)$$ The intermediate value theorems gives us c such that $$ g(c)=0$$ and the theorem follows.