设 $$\bex k\geq 2,\quad f\in C^k(\bbR),\quad M_j=\sup_{x\in\bbR}|f^{(j)}(x)|\ (j=0,1,\cdots,k). \eex$$ 则 $$\bex M_j\leq 2^\frac{j(k-j)}{2}M_0^{1-\frac{j}{k}}M_k^\frac{j}{k}\ (j=0,1,\cdots,k). \eex$$

 

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 $$\bex f(x+h)=f(x)+f'(x)h+\frac{f''(\xi)}{2}h^2, \eex$$ $$\bex f(x-h)=f(x)-f'(x)h+\frac{f''(\eta)}{2}h^2. \eex$$ 相减而有 $$\bex |f'(x)|\cdot 2h \leq 2M_0+M_2h^2\ra |f'(x)|\leq \frac{M_0}{h}+\frac{M_2h}{2}. \eex$$ 取 $$\bex h=\sqrt{\frac{2M_0}{M_2}}\lra \frac{M_0}{h}=\frac{M_2h}{2}, \eex$$ 则 $$\bex |f'(x)|\leq 2\cdot \frac{M_2}{2}\sqrt{\frac{2M_0}{M_2}}=\sqrt{2M_0M_2}. \eex$$ 假设结论当 $k=n$ 时成立, 则当 $k=n+1$ 时, 对 $0<j<n+1$, $$\beex \bea M_j&\leq 2^\frac{j(n-j)}{2}M_0^{1-\frac{j}{n}}M_n^\frac{j}{n}\\ &\leq 2^\frac{j(n-j)}{2}M_0^{1-\frac{j}{n}}\sex{2^\frac{n}{2} M_0^{1-\frac{n}{n+1}}M_{n+1}^\frac{n}{n+1}}^\frac{j}{n}\\ &=2^\frac{j(n+1-j)}{2} M_0^{1-\frac{j}{n+1}}M_{n+1}^\frac{j}{n+1}. \eea \eeex$$