Universal chord theorem

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

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]

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 ${\displaystyle f(x)=f(x+1/2)}$.

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 ${\displaystyle n=2}$.

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.

Let ${\displaystyle r\in \mathbb {R} }$ be arbitrary, and consider the function ${\displaystyle f:[0,1]\to \mathbb {R} }$ defined by ${\displaystyle f(x)=\sin ^{2}\left({\frac {\pi x}{r}}\right)-x\sin ^{2}\left({\frac {\pi }{r}}\right)}$. It is immediate that ${\displaystyle f}$ is continuous, and ${\displaystyle f(0)=f(1)=0}$. If some ${\displaystyle x\in [0,1]}$ satisfies ${\displaystyle f(x)=f(x+r)}$, then ${\displaystyle r\sin ^{2}\left({\frac {\pi }{r}}\right)=0}$ which implies that ${\displaystyle r={\frac {1}{n}}}$ for some integer ${\displaystyle n}$. Therefore the theorem does not hold for non-integer values of ${\displaystyle n}$. ^{[citation needed]}