Question

设f(x)在[0,1]上连续,在(0,1)内可导,且f(0)=f(1)=0 ,记M=max{ l f(x) l ,x∈[

Answer 1

由f(x)在[0,1]上连续,|f(x)|在[0,1]上可以取得最大值M.

设在c∈[0,1]处|f(c)| = M.

若c ≥ 1/2,在[c,1]上由Lagrange中值定理,存在ζ∈(c,1)使f'(ζ) = (f(1)-f(c))/(1-c) = -f(c)/(1-c).

此时|f'(ζ)| = |f(c)/(1-c)| = M/(1-c) ≥ 2M.

若c ≤ 1/2,在[0,c]上由Lagrange中值定理,存在ζ∈(0,c)使f'(ζ) = (f(c)-f(0))/(c-0) = f(c)/c.

此时|f'(ζ)| = |f(c)/c| = M/c ≥ 2M.