真神问的积分不等式

陈洪葛 posted @ Jun 10, 2014 08:29:07 AM in 数学分析 , 1351 阅读

设$f(x)$在$[0,\pi]$上可微,且满足$(f'(x))^2 \in R[a,b]$,若$\displaystyle \int_{0}^{\pi}f(x)dx=0$,则
\[ \int_{0}^{\pi}f^{2}(x)dx\leq \int_{0}^{\pi}(f'(x))^2 dx \]


证明:我们把$f(x)$以$\pi$为周期先偶延拓到$[-\pi,\pi]$,再以$2\pi$为周期延拓到$\mathbb{R}$,这时$2k\pi,k\in \mathbf{N}^{+}$是$f(x)$的周期,注意到
\[ f(x+2\pi)=f(x),f(x)=f(-x) \]
两边对$x$求导,得到
\[ f'(x+2\pi)=f'(x),f'(x)=-f'(-x) \]
就是说$2k\pi,k\in \mathbf{N}^{+}$也是$f'(x)$的周期,而
\[ \int_{0}^{\pi}f^{2}(x)dx=\int_{-\pi}^{0}f^{2}(x)dx=\int_{\pi}^{2\pi}f^2(x)dx \]

\[ \int_{0}^{\pi}(f'(x))^2dx=\int_{-\pi}^{0}(f'(x))^2dx=\int_{\pi}^{2\pi}(f'(x))^2dx \]
不等式显然等价于
\[ \int_{0}^{2\pi}f^{2}(x)dx\leq \int_{0}^{2\pi}(f'(x))^2dx \]
设$f(0)=f(2\pi)=a$,这时,注意到
\[ 0\leq (f(x)-a)^2\cot{x}\leq \left(\int_{0}^{x}f'(t)dt\right)^{2}\frac{1}{\sin{x}}\leq \int_{0}^{x}dt\int_{0}^{x}(f'(t))^{2}dt\frac{1}{\sin{x}}=\frac{x}{\sin{x}}\int_{0}^{x}(f'(t))^{2}dt\to 0 (x\to 0^{+}) \]

\[ \lim_{x\to 0}(f(x)-a)^{2}\cot x=0 \]
同理也有
\[ \lim_{x\to 2\pi}(f(x)-a)^{2}\cot x=0 \]
不妨延拓定义
\[ (f(x)-a)^{2}\cot x\bigg|_{x=0}=(f(x)-a)^{2}\cot x\bigg|_{x=2\pi}=0 \]
现在,我们有
\begin{align*}
&\int_{0}^{2\pi}\bigg((f'(x))^2-(f(x)-a)^2-[f'(x)-(f(x)-a)\cot x]^{2}\bigg)dx\\
&=\int_{0}^{2\pi}\bigg(-(f(x)-a)^2\csc^{2}x+2f'(x)(f(x)-a)\cot x\bigg)dx\\
&=(f(x)-a)^{2}\cot x\bigg|_{0}^{2\pi}\\
&=0
\end{align*}

\[ \int_{0}^{2\pi}[(f'(x))^2-(f(x)-a)^2]dx=\int_{0}^{2\pi}[f'(x)-(f(x)-a)\cot x]^{2}dx\geq 0 \]
而注意到
\[ \int_{0}^{2\pi}f(x)dx=0 \]则
\[ \int_{0}^{2\pi}[(f'(x))^2-(f(x)-a)^2]dx=\int_{0}^{2\pi}((f'(x))^2-(f(x))^2)dx-2\pi a^2\geq 0 \]
就是说
\[ \int_{0}^{2\pi}(f'(x))^2dx\geq \int_{0}^{2\pi}(f(x))^2dx+2\pi a^2\geq \int_{0}^{2\pi}(f(x))^2dx \]
Hence we are done!


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