Question

一道可能是关于中值问题的证明题f（x）在[a,b]上有三阶连续导数,f（a）=f（b）=0,证明：在[a,b]上,max

Answer 1

用带Lagrange余项的Taylor展开,广义上也算是中值定理吧.

首先,由f(x)在[a,b]连续,可设|f(x)|在c处取得[a,b]上的最大值.

若c = a或c = b,有f(x)在[a,b]上恒等于0,不等式显然成立.

若a < c < b,可得f'(c) = 0 (c是f(x)的极值点).

考虑f(x)在c处的Taylor展开(Lagrange余项):

对x ∈ [a,b],在x与c之间存在t,使f(x) = f(c)+f'(c)(x-c)+f"(t)(x-c)²/2 = f(c)+f"(t)(x-c)²/2.

当c ≤ (a+b)/2,在上式中取x = a得0 = f(a) = f(c)+f'(c)(a-c)+f"(t)(a-c)²/2 = f(c)+f"(t)(a-c)²/2.

故|f(c)| = |f"(t)|(a-c)²/2 ≤ |f"(t)|(a-(a+b)/2)²/2 = |f"(t)|(a-b)²/8 ≤ (a-b)²/8·max|f"(x)|.

而当c ≥ (a+b)/2,改取x = b同样可得|f(c)| ≤ (a-b)²/8·max|f"(x)|.

由假设,|f(c)| = max|f(x)|,即得max|f(x)| ≤ (a-b)²/8·max|f"(x)|,不等式成立.