流行积分不等式

流行积分不等式

并不是有多流行，只是这个不等式的名字就叫流行不等式

设 $f(x)$ 在 $[a,b]$ 上可微，且当 $ x \in(a,b)$ 时，$ \displaystyle 0 < f’(x) < \frac{2}{n+1}$，$f(a)=0$，则有 $\displaystyle \left[\int_{a}^{b}f^n(x), \mathrm{d}x\right]^2>\int_{a}^{b}f^{2n+1}(x), \mathrm{d}x$

证明：

法一：直接构造变限积分函数求导

令
$$
F(x)=\left[\int_{a}^{x}f^n(t), \mathrm{d}t\right]^2-\int_{a}^{x}f^{2n+1}(t), \mathrm{d}t
$$
要证 $ F(b)>0 $，由 $ F(a)=0 $，即证 $ F^\prime(x)>0 $
$$
\begin{align*}
F^\prime(x)&=2f^n(x)\int_{a}^{x}f^n(t), \mathrm{d}t - f^{2n+1}(x)\
&= f^n(x)\left[2\int_{a}^{x}f^n(t), \mathrm{d}t - f^{n+1}(x)\right]
\end{align*}
$$
令
$$
g(x)=2\int_{a}^{x}f^n(t), \mathrm{d}t - f^{n+1}(x)
$$

$$
\begin{align*}
g^\prime(x)&=2f^n(x)-(n+1)f^n(x)f^\prime(x)\
&=f^n(x)\left[2-(n+1)f^\prime(x)\right]
\end{align*}
$$

由 $ f(a)=0 $，$ f^\prime(x)>0 $，故 $ f(x)>0 $，$ f^n(x)>0 $

且
$$
0<f’\left(x\right)<\frac{2}{n+1}
$$
故
$$
\begin{align*}
g^\prime(x)&>0 \
g(x)>g&(a)=0
\end{align*}
$$
则
$$
\begin{align*}
F^\prime(x)&>0\
\Rightarrow F(x)>F&(a)=0
\end{align*}
$$
即
$$
\left[\int_{a}^{b}f^n(x), \mathrm{d}x\right]^2>\int_{a}^{b}f^{2n+1}(x), \mathrm{d}x
$$
法二：柯西中值定理

由 $ f(a)=0 $，$ f^\prime(x)>0 $，$ \Rightarrow f(x) $ 在 $(a,b)$ 上恒正

要证
$$
\left[\int_{a}^{b}f^n(x), \mathrm{d}x\right]^2>\int_{a}^{b}f^{2n+1}(x), \mathrm{d}x
$$
即证
$$
\frac{\displaystyle \left[\int_{a}^{b}f^n(x), \mathrm{d}x\right]^2}{\displaystyle \int_{a}^{b}f^{2n+1}(x), \mathrm{d}x}>1
$$
令
$$
\begin{align*}
\displaystyle F(x)&=\left[\int_{a}^{x}f^n(t), \mathrm{d}t\right]^2,\quad F(a)=0\
\displaystyle G(x)&=\int_{a}^{x}f^{2n+1}(t), \mathrm{d}t,\quad G(a)=0
\end{align*}
$$
则原式可写成：
$$
\frac{\displaystyle \left[\int_{a}^{b}f^n(x), \mathrm{d}x\right]^2 - 0}{\displaystyle \int_{a}^{b}f^{2n+1}(x), \mathrm{d}x - 0}=\frac{F(b)-F(a)}{G(b)-G(a)}
$$
由柯西中值定理得：
$$
\frac{\displaystyle \left[\int_{a}^{b}f^n(x), \mathrm{d}x\right]^2}{\displaystyle \int_{a}^{b}f^{2n+1}(x), \mathrm{d}x}=\frac{F^\prime(\xi)}{G^\prime(\xi)}=\frac{\displaystyle 2f^n(\xi)\int_{a}^{\xi}f^n(t), \mathrm{d}t}{f^{2n+1}(\xi)}
$$
其中 $\xi \in \left(a,b\right)$

令
$$
H(x)=\int_{a}^{x}f^n(t), \mathrm{d}t,\quad H(a)=0
$$
则
$$
\begin{align*}
\frac{F^\prime(\xi)}{G^\prime(\xi)}&=\frac{2(H(\xi)-H(a))}{f^{n+1}(\xi)-f^{n+1}(a)}=\frac{2H^\prime(\eta)}{(n+1)f^n(\eta)f^\prime(\eta)}\
&=\frac{2f^n(\eta)}{(n+1)f^n(\eta)f^\prime(\eta)}\
&=\frac{2}{(n+1)f^\prime(\eta)}>1
\end{align*}
$$
其中 $\eta\in\left(a,\xi\right)$

即
$$
\frac{\displaystyle \left[\int_{a}^{b}f^n(x), \mathrm{d}x\right]^2}{\displaystyle \int_{a}^{b}f^{2n+1}(x), \mathrm{d}x}>1
$$

$$
\left[\int_{a}^{b}f^n(x), \mathrm{d}x\right]^2>\int_{a}^{b}f^{2n+1}(x), \mathrm{d}x
$$

下面看一个具体例子

设 $f(x)$ 在 $[0,1]$ 可微，且当 $x \in(0,1)$ 时，$0 < f’(x) < 1$，$f(0)=0$

证明：$\displaystyle \left[\int_{0}^{1}f(x), \mathrm{d}x\right]^2>\int_{0}^{1}f^3(x), \mathrm{d}x$

证明：

法一：

令
$$
F(x)=\left[\int_{0}^{x}f(t), \mathrm{d}t\right]^2-\int_{0}^{x}f^3(t), \mathrm{d}t,\quad F(0)=0
$$
即证 $ F(1)>0 $
$$
\begin{align*}
F^\prime(x)&=2f(x)\cdot\int_{0}^{x}f(t), \mathrm{d}t-f^3(x)\
& = f(x)\left[2\int_{0}^{x}f(t), \mathrm{d}t - f^2(x)\right]
\end{align*}
$$
令
$$
G(x)=2\int_{0}^{x}f(t), \mathrm{d}t - f^2(x),\quad G(0)=0
$$

$$
\begin{align*}
G^\prime(x)&=2f(x)-2f(x)f^\prime(x)\
&=2f(x)[1 - f^\prime(x)]>0
\end{align*}
$$

故 $ G(x)>G(0)=0 $

由 $ f(0)=0 $，$ f^\prime(x)>0 $，故 $ f(x)>0 $ 在 $ (0,1) $ 上恒成立

则
$$
F^\prime(x)=f(x)G(x)>0
$$
故
$$
\begin{align*}
F(x)&>F(0)=0\
\Rightarrow F(1)=&\left[\int_{0}^{1}f(x), \mathrm{d}x\right]^2-\int_{0}^{1}f^3(x), \mathrm{d}x>0\
\end{align*}
$$

$$
\left[\int_{0}^{1}f(x), \mathrm{d}x\right]^2>\int_{0}^{1}f^3(x), \mathrm{d}x
$$

法二：

令
$$
\begin{align*}
F(x)&=\left[\int_{0}^{x}f(t), \mathrm{d}t\right]^2,\quad F(0)=0\
G(x)&=\int_{0}^{x}f^3(t), \mathrm{d}t,\quad G(0)=0
\end{align*}
$$
则
$$
\begin{align*}
\frac{\displaystyle \left[\int_{0}^{1}f(t), \mathrm{d}t\right]^2}{\displaystyle \int_{0}^{1}f^3(t), \mathrm{d}t}&=\frac{\displaystyle \left[\int_{0}^{1}f(t), \mathrm{d}t\right]^2 - 0}{\displaystyle \int_{0}^{1}f^3(t), \mathrm{d}t - 0}\
&=\frac{F(1)-F(0)}{G(1)-G(0)}\
&=\frac{F^\prime(\xi)}{G^\prime(\xi)}\
&=\frac{\displaystyle 2f(\xi)\cdot\int_{0}^{\xi}f(t), \mathrm{d}t}{f^3(\xi)}\
&=\frac{\displaystyle 2\left(\int_{0}^{\xi}f(t), \mathrm{d}t - 0\right)}{f^2(\xi)-0}\
&=\frac{2f(\eta)}{2f(\eta)f^\prime(\eta)}\
&=\frac{1}{f^\prime(\eta)}>1
\end{align*}
$$
其中 $\xi \in \left(a,b\right),\eta\in\left(0,\xi\right)$

即
$$
\frac{\displaystyle \left[\int_{0}^{1}f(t), \mathrm{d}t\right]^2}{\displaystyle \int_{0}^{1}f^3(t), \mathrm{d}t}>1
$$

$$
\Rightarrow\left[\int_{0}^{1}f(x), \mathrm{d}x\right]^2>\int_{0}^{1}f^3(x), \mathrm{d}x
$$