Question

大一高数题：设f(x)在闭区间[0,1]上连续,f(0)=0,f(1)=1,证明:存在ξ∈(0,1),使得f(ξ-1/3)=f(ξ)-1/3.

Answer 1

令g(x)=f(x)-f(x-1/3)-1/3,
则g(x)连续；
设m=<g(x)<=M;
g(1)=f(1)-f(2/3)-1/3=2/3-f(2/3);
g(2/3)=f(2/3)-f(1/3)-1/3;
g(1/3)=f(1/3)-1/3;
g(1)+g(2/3)+g(1/3)=0;
又因为3m=<g(1)+g(2/3)+g(1/3))<=3M；
所以m<=0且M>=0
所以存在ξ∈(0,1),使得g(ξ)=0,故f(ξ-1/3)=f(ξ)-1/3.

Answer 2

设F(x)=f(x-1/3)-f(x)+1/3
F(1/3)=f(0)-f(1/3)+1/3=-f(1/3)+1/3
F(2/3)=f(1/3)-f(2/3)+1/3
F(1)=f(2/3)-f(1)+1/3=f(2/3)-2/3

F(1/3)+F(2/3)=-f(2/3)+2/3 ,由介值性定理，至少存在a,(1/3《a《2/3),使：
F(a)=(F(1/3)+F(2/3))/2=(-f(2/3)+2/3)/2
故：F(a)F(1)=(-f(2/3)+2/3)/2*(f(2/3)-2/3)《0, 由根的存在性定理：
至少存在ξ，使得F(ξ)=0 ,即： f(ξ-1/3)=f(ξ)-1/3