In mathematical analysis, the universal chord theorem states that if a function f is continuous on [a,b] and satisfies f ( a ) = f ( b ) {\displaystyle f(a)=f(b)} , then for every natural number n {\displaystyle n} , there exists some x ∈ [ a , b ] {\displaystyle x\in [a,b]} such that f ( x ) = f ( x + b − a n ) {\displaystyle f(x)=f\left(x+{\frac {b-a}{n}}\right)} .
History The theorem was published by Paul Lévy in 1934 as a generalization of Rolle's theorem.
Statement of the theorem Let H ( f ) = { h ∈ [ 0 , + ∞ ) : f ( x ) = f ( x + h ) for some x } {\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 h ∈ H ( f ) {\displaystyle h\in H(f)} , then h n ∈ H ( f ) {\displaystyle {\frac {h}{n}}\in H(f)}
for all natural numbers n.
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 f ( x ) {\displaystyle f(x)} is continuous on some interval I = [ a , b ] {\displaystyle I=[a,b]} with the condition that f ( a ) = f ( b ) {\displaystyle f(a)=f(b)} , then there exists some x ∈ [ a , b ] {\displaystyle x\in [a,b]} such that f ( x ) = f ( x + b − a 2 ) {\displaystyle f(x)=f\left(x+{\frac {b-a}{2}}\right)} . In less generality, if f : [ 0 , 1 ] → R {\displaystyle f:[0,1]\rightarrow \mathbb {R} } is continuous and f ( 0 ) = f ( 1 ) {\displaystyle f(0)=f(1)} , then there exists x ∈ [ 0 , 1 2 ] {\displaystyle x\in \left[0,{\frac {1}{2}}\right]} that satisfies f ( x ) = f ( x + 1 / 2 ) {\displaystyle f(x)=f(x+1/2)} .
Proof of n = 2 Consider the function g : [ a , b + a 2 ] → R {\displaystyle g:\left[a,{\dfrac {b+a}{2}}\right]\to \mathbb {R} } defined by g ( x ) = f ( x + b − a 2 ) − f ( x ) {\displaystyle g(x)=f\left(x+{\dfrac {b-a}{2}}\right)-f(x)} . Being the sum of two continuous functions, g {\displaystyle g} is continuous, g ( a ) + g ( b + a 2 ) = f ( b ) − f ( a ) = 0 {\displaystyle g(a)+g\left({\dfrac {b+a}{2}}\right)=f(b)-f(a)=0} . It follows that g ( a ) ⋅ g ( b + a 2 ) ≤ 0 {\displaystyle g(a)\cdot g\left({\dfrac {b+a}{2}}\right)\leq 0} and by applying the intermediate value theorem, there exists c ∈ [ a , b + a 2 ] {\displaystyle c\in \left[a,{\dfrac {b+a}{2}}\right]} such that g ( c ) = 0 {\displaystyle g(c)=0} , so that f ( c ) = f ( c + b − a 2 ) {\displaystyle f(c)=f\left(c+{\dfrac {b-a}{2}}\right)} . This concludes the proof of the theorem for n = 2 {\displaystyle n=2} .
Proof of general case The proof of the theorem in the general case is very similar to the proof for n = 2 {\displaystyle n=2}
Let n {\displaystyle n} be a non negative integer, and consider the function g : [ a , b − b − a n ] → R {\displaystyle g:\left[a,b-{\dfrac {b-a}{n}}\right]\to \mathbb {R} } defined by g ( x ) = f ( x + b − a n ) − f ( x ) {\displaystyle g(x)=f\left(x+{\dfrac {b-a}{n}}\right)-f(x)} . Being the sum of two continuous functions, g {\displaystyle g} is continuous. Furthermore, ∑ k = 0 n − 1 g ( a + k ⋅ b − a n ) = 0 {\displaystyle \sum _{k=0}^{n-1}g\left(a+k\cdot {\dfrac {b-a}{n}}\right)=0} . It follows that there exists integers i , j {\displaystyle i,j} such that g ( a + i ⋅ b − a n ) ≤ 0 ≤ g ( a + j ⋅ b − a n ) {\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 g ( c ) = 0 {\displaystyle g(c)=0} and the theorem follows.
Counterexample for non-integer n Let r ∈ R {\displaystyle r\in \mathbb {R} } be arbitrary, and consider the function f : [ 0 , 1 ] → R {\displaystyle f:[0,1]\to \mathbb {R} } defined by f ( x ) = sin 2 ( π x r ) − x sin 2 ( π r ) {\displaystyle f(x)=\sin ^{2}\left({\frac {\pi x}{r}}\right)-x\sin ^{2}\left({\frac {\pi }{r}}\right)} . It is immediate that f {\displaystyle f} is continuous, and f ( 0 ) = f ( 1 ) = 0 {\displaystyle f(0)=f(1)=0} . If some x ∈ [ 0 , 1 ] {\displaystyle x\in [0,1]} satisfies f ( x ) = f ( x + r ) {\displaystyle f(x)=f(x+r)} , then r sin 2 ( π r ) = 0 {\displaystyle r\sin ^{2}\left({\frac {\pi }{r}}\right)=0} which implies that r = 1 n {\displaystyle r={\frac {1}{n}}} for some integer n {\displaystyle n} . Therefore the theorem does not hold for non-integer values of n {\displaystyle n} .
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