设 $$ex kgeq 2,quad fin C^k(bR),quad M_j=sup_{xinbR}|f^{(j)}(x)| (j=0,1,cdots,k). eex$$ 则 $$ex M_jleq 2^frac{j(k-j)}{2}M_0^{1-frac{j}{k}}M_k^frac{j}{k} (j=0,1,cdots,k). eex$$
证明:
(1). 仅需对 $0<j<k$ 证明结论成立.
(2). 往对 $k$ 作数学归纳法. 当 $k=2$ 时, 对 $j=1$, 由 $$ex f(x+h)=f(x)+f'(x)h+frac{f''(xi)}{2}h^2, eex$$ $$ex f(x-h)=f(x)-f'(x)h+frac{f''(eta)}{2}h^2. eex$$ 相减而有 $$ex |f'(x)|cdot 2h leq 2M_0+M_2h^2 a |f'(x)|leq frac{M_0}{h}+frac{M_2h}{2}. eex$$ 取 $$ex h=sqrt{frac{2M_0}{M_2}}lra frac{M_0}{h}=frac{M_2h}{2}, eex$$ 则 $$ex |f'(x)|leq 2cdot frac{M_2}{2}sqrt{frac{2M_0}{M_2}}=sqrt{2M_0M_2}. eex$$ 假设结论当 $k=n$ 时成立, 则当 $k=n+1$ 时, 对 $0<j<n+1$, $$eex ea M_j&leq 2^frac{j(n-j)}{2}M_0^{1-frac{j}{n}}M_n^frac{j}{n}\ &leq 2^frac{j(n-j)}{2}M_0^{1-frac{j}{n}}sex{2^frac{n}{2} M_0^{1-frac{n}{n+1}}M_{n+1}^frac{n}{n+1}}^frac{j}{n}\ &=2^frac{j(n+1-j)}{2} M_0^{1-frac{j}{n+1}}M_{n+1}^frac{j}{n+1}. eea eeex$$