证明:将m改为x,设F(x) = f(x)-f(x+1/n)
因为f(x)在[0,1]上连续,所以F(x)在[0,1]上连续,得:
F(0) = f(0)-f(1/n)
F(1/n) = f(1/n)-f(2/n)
F(2/n) = f(2/n)-f(3/n)
……
F(n-2/n) = f(n-2/n)-f(n-1/n)
F(n-1/n) = f(n-1/n)-f(1)
将上面n个式子相加,得:F(0)+F(1/n)+……+F(n-1/n)=f(1)-f(0)=0(*)
所以由(*)式可知:必存在a,b属于(0,1),使得F(a)*F(b)