In mathematical analysis, the universal chord theorem states that if a function f is continuous on [''a'',''b''] and satisfies
f(a)=f(b)
n
x\in[a,b]
f(x)=f\left(x+
b-a | |
n |
\right)
The theorem was published by Paul Lévy in 1934 as a generalization of Rolle's Theorem.[2]
Let
H(f)=\{h\in[0,+infty):f(x)=f(x+h)forsomex\}
h\inH(f)
h | |
n |
\inH(f)
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)
I=[a,b]
f(a)=f(b)
x\in[a,b]
f(x)=f\left(x+
b-a | |
2 |
\right)
In less generality, if
f:[0,1] → \R
f(0)=f(1)
x\in\left[0,
1 | |
2 |
\right]
f(x)=f(x+1/2)
Consider the function
g:\left[a,\dfrac{b+a}{2}\right]\toR
g(x)=f\left(x+\dfrac{b-a}{2}\right)-f(x)
g
g(a)+g(\dfrac{b+a}{2})=f(b)-f(a)=0
g(a) ⋅ g(\dfrac{b+a}{2})\le0
c\in\left[a,\dfrac{b+a}{2}\right]
g(c)=0
f(c)=f\left(c+\dfrac{b-a}{2}\right)
n=2
The proof of the theorem in the general case is very similar to the proof for
n=2
n
g:\left[a,b-\dfrac{b-a}{n}\right]\toR
g(x)=f\left(x+\dfrac{b-a}{n}\right)-f(x)
g
n-1 | |
\sum | |
k=0 |
g\left(a+k ⋅ \dfrac{b-a}{n}\right)=0
i,j
g\left(a+i ⋅ \dfrac{b-a}{n}\right)\le0\leg\left(a+j ⋅ \dfrac{b-a}{n}\right)
g(c)=0