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Universal chord theorem

Guarantees chords of length 1/n exist for functions satisfying certain conditions

Inmathematical 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]

History[edit]

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

Statement of the theorem[edit]

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[edit]

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 ${\displaystyle f:[0,1]\rightarrow \mathbb {R} }$iscontinuous 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[edit]

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({\dfrac {b+a}{2}})=f(b)-f(a)=0} . It follows that ${\displaystyle g($a)\cdot g({\dfrac {b+a}{2}})\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)} . Which concludes the proof of the theorem for ${\displaystyle n=2}$

Proof of general case[edit]

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.

See also[edit]

• Intermediate value theorem
• Borsuk–Ulam theorem
• Rolle's theorem

References[edit]