Universal chord theorem

The theorem was published by Paul Lévy in 1934 as a generalization of Rolle's theorem.Paul Levy, "Sur une Généralisation du Théorème de Rolle", C. R. Acad. Sci., Paris, 198 (1934) 424–425.

Let $H(f)\; =\; \backslash \{\; h\; \backslash in\; [0,\; +\backslash infty)\; :\; f(x)\; =\; f(x+h)\; \backslash text\{\; for\; some\; \}\; x\; \backslash \}$ denote the chord set of the function f. If f is a continuous function and $h\; \backslash in\; H(f)$, then $\backslash frac\; h\; n\; \backslash in\; H(f)$

for all natural numbers n.

{{cite journal|last1=Oxtoby|first1=J.C.|title=Horizontal Chord Theorems|journal=The American Mathematical Monthly|date=May 1978|volume=79|pages=468–475|doi=10.2307/2317564}}

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 $x\; \backslash in\; [a,b]$ such that $f(x)\; =\; f\backslash left(x\; +\; \backslash frac\{b-a\}\{2\}\backslash right)$.

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

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

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 $g:\backslash left[a,\; b\; -\; \backslash dfrac\{b-a\}\{n\}\backslash right]\backslash to\backslash mathbb\{R\}$ defined by $g(x)\; =\; f\backslash left(x\; +\; \backslash dfrac\{b-a\}\{n\}\backslash right)\; -\; f(x)$. Being the sum of two continuous functions, $g$ is continuous. Furthermore, $\backslash sum\_\{k=0\}^\{n-1\}g\backslash left(a+k\backslash cdot\backslash dfrac\{b-a\}\{n\}\backslash right)\; =\; 0$. It follows that there exists integers $i,j$ such that $g\backslash left(a+i\backslash cdot\backslash dfrac\{b-a\}\{n\}\backslash right)\backslash le\; 0\backslash le\; g\backslash left(a+j\backslash cdot\backslash dfrac\{b-a\}\{n\}\backslash right)$

The intermediate value theorems gives us c such that $g(c)=0$ and the theorem follows.

Category:Theorems in mathematical analysis