若函数f(x)在[0,1]上连续,f(0)=f(1),则对任意自然数n,存在ξ∈[0,1],使得f(ξ+1/n)=f(ξ

设F(x)=f(x+1/n)-f(x)
F(0)=f(1/n)-f(0)
F(1/n)=f(2/n)-f(1/n)
…
F[(n-1)/n]=f(1)-f[(n-1)/n]
那么F(0)+F(1/n)+…+F[(n-1)/n]
=f(1/n)-f(0)+f(2/n)-f(1/n)+…+f(1)-f[(n-1)/n]
=f(1)-f(0)
=0
所以F(0)=F(1/n)=…F[(n-1)/n]=0或存在F(i/n)和F(j/n)符号相反（0≤i