好题00014- $\displaystyle\sup_{r\in (0,R)}\left|\frac{f'(r)}{r}\right|\leq \frac{1}{n}\sup_{r\in (0,R)}\left|f''(r)+\frac{n-1}{r}f'(r)\right|.$

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设 $\displaystyle n \geq 2$ 为整数, 函数 $\displaystyle f \in {C}^{2}\left( \left\lbrack {0, R}\right\rbrack \right)$ 满足 $\displaystyle {f}_{r }\left( 0\right) = 0$ . 试证明
$\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} \sup_{r\in (0,R)}\left|\frac{f'(r)}{r}\right|\leq \frac{1}{n}\sup_{r\in (0,R)}\left|f''(r)+\frac{n-1}{r}f'(r)\right|.\tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

[张祖锦注: $\displaystyle f_r(0)=0$ 可以去掉!] 来源:  link

纸质资料/答疑/pdf1/pdf2 /
对 $\displaystyle \forall\ 0\lt r\lt R$ ,
$\displaystyle \tiny\boxed{@跟锦数学微信公众号}$

$$\begin{aligned} &\left|\frac{f'(r)}{r}\right| =\left|\frac{r^{n-1}f'(r)}{r^n}\right| =\left|\frac{\rho^{n-1}f'(\rho)|_{\rho=0}^{\rho=r}}{r^n}\right| \overset{\tiny\mbox{中值}}{\tiny\mbox{Cauchy}}\left|\frac{[\rho^{n-1}f'(\rho)]'|_{\rho=\xi}}{n\xi^{n-1}}\right|\\ =&\left|\frac{(n-1)\xi^{n-2}f'(\xi)+\xi^{n-1}f''(\xi)}{n\xi^{n-1}}\right| =\frac{1}{n}\left|f''(\xi)+\frac{n-1}{\xi}f'(\xi)\right| \leq \mbox{右端}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$

跟锦数学微信公众号. 在线资料/公众号/

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