Overview
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\left(\dfrac b+a 2 \right) = f(b) - f(a) = 0 . It follows that g(a)\cdot g\left(\dfrac b+a 2 \right)\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) . 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)\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.
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 .