Universal chord theorem
In mathematical analysis, the universal chord theorem states that if a function f is continuous on [a,b] and satisfies [math]\displaystyle{ f(a) = f(b) }[/math], then for every natural number [math]\displaystyle{ n }[/math], there exists some [math]\displaystyle{ x \in [a,b] }[/math] such that [math]\displaystyle{ f(x) = f\left(x + \frac{b-a}{n}\right) }[/math].[1]
History
The theorem was published by Paul Lévy in 1934 as a generalization of Rolle's Theorem.[2]
Statement of the theorem
Let [math]\displaystyle{ H(f) = \{ h \in [0, +\infty) : f(x) = f(x+h) \text{ for some } x \} }[/math] denote the chord set of the function f. If f is a continuous function and [math]\displaystyle{ h \in H(f) }[/math], then [math]\displaystyle{ \frac h n \in H(f) }[/math] 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 [math]\displaystyle{ f(x) }[/math] is continuous on some interval [math]\displaystyle{ I = [a,b] }[/math] with the condition that [math]\displaystyle{ f(a) = f(b) }[/math], then there exists some [math]\displaystyle{ x \in [a,b] }[/math] such that [math]\displaystyle{ f(x) = f\left(x + \frac{b-a}{2}\right) }[/math].
In less generality, if [math]\displaystyle{ f : [0,1] \rightarrow \R }[/math] is continuous and [math]\displaystyle{ f(0) = f(1) }[/math], then there exists [math]\displaystyle{ x \in \left[0,\frac{1}{2}\right] }[/math] that satisfies [math]\displaystyle{ f(x) = f(x+1/2) }[/math].
Proof of n = 2
Consider the function [math]\displaystyle{ g:\left[a, \dfrac{b+a}{2}\right]\to\mathbb{R} }[/math] defined by [math]\displaystyle{ g(x) = f\left(x+\dfrac{b-a}{2}\right) - f(x) }[/math]. Being the sum of two continuous functions, [math]\displaystyle{ g }[/math] is continuous, [math]\displaystyle{ g(a) + g(\dfrac{b+a}{2}) = f(b) - f(a) = 0 }[/math]. It follows that [math]\displaystyle{ g(a)\cdot g(\dfrac{b+a}{2})\le 0 }[/math] and by applying the intermediate value theorem, there exists [math]\displaystyle{ c\in \left[a, \dfrac{b+a}{2}\right] }[/math] such that [math]\displaystyle{ g(c) = 0 }[/math], so that [math]\displaystyle{ f(c) = f\left(c + \dfrac{b-a}{2}\right) }[/math]. Which concludes the proof of the theorem for [math]\displaystyle{ n = 2 }[/math]
Proof of general case
The proof of the theorem in the general case is very similar to the proof for [math]\displaystyle{ n = 2 }[/math] Let [math]\displaystyle{ n }[/math] be a non negative integer, and consider the function [math]\displaystyle{ g:\left[a, b - \dfrac{b-a}{n}\right]\to\mathbb{R} }[/math] defined by [math]\displaystyle{ g(x) = f\left(x + \dfrac{b-a}{n}\right) - f(x) }[/math]. Being the sum of two continuous functions, [math]\displaystyle{ g }[/math] is continuous. Furthermore, [math]\displaystyle{ \sum_{k=0}^{n-1}g\left(a+k\cdot\dfrac{b-a}{n}\right) = 0 }[/math]. It follows that there exists integers [math]\displaystyle{ i,j }[/math] such that [math]\displaystyle{ g\left(a+i\cdot\dfrac{b-a}{n}\right)\le 0\le g\left(a+j\cdot\dfrac{b-a}{n}\right) }[/math] The intermediate value theorems gives us c such that [math]\displaystyle{ g(c)=0 }[/math] and the theorem follows.
See also
References
- ↑ Rosenbaum, J. T. (May, 1971) The American Mathematical Monthly, Vol. 78, No. 5, pp. 509–513
- ↑ Paul Levy, "Sur une Généralisation du Théorème de Rolle", C. R. Acad. Sci., Paris, 198 (1934) 424–425.
- ↑ Oxtoby, J.C. (May 1978). "Horizontal Chord Theorems". The American Mathematical Monthly 79: 468–475. doi:10.2307/2317564.
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