Universal chord theorem

In mathematical analysis, the universal chord theorem states that if a function f is continuous on [a,b] and satisfies ${\displaystyle f(a)=f(b)}$, then for every natural number ${\displaystyle n}$, there exists some ${\displaystyle x\in [a,b]}$ such that ${\displaystyle f(x)=f\left(x+{\frac {b-a}{n}}\right)}$.^[1]

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.

History

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

Statement of the theorem

Let ${\displaystyle H(f)=\{h\in [0,+\infty ):f(x)=f(x+h){\text{ for some }}x\}}$  denote the chord set of the function f. If f is a continuous function and ${\displaystyle h\in H(f)}$ , then ${\displaystyle {\frac {h}{n}}\in H(f)}$  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 ${\displaystyle f(x)}$  is continuous on some interval ${\displaystyle I=[a,b]}$  with the condition that ${\displaystyle f(a)=f(b)}$ , then there exists some ${\displaystyle x\in [a,b]}$  such that ${\displaystyle f(x)=f\left(x+{\frac {b-a}{2}}\right)}$ .

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

Proof of n = 2

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

Proof of general case

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

References

1. Rosenbaum, J. T. (May, 1971) The American Mathematical Monthly, Vol. 78, No. 5, pp. 509–513
2. Paul Levy, "Sur une Généralisation du Théorème de Rolle", C. R. Acad. Sci., Paris, 198 (1934) 424–425.
3. Oxtoby, J.C. (May 1978). "Horizontal Chord Theorems". The American Mathematical Monthly. 79: 468–475. doi:10.2307/2317564.