来自考研教辅书上的一个积分不等式

陈洪葛 posted @ May 05, 2014 05:42:41 PM in 数学分析 , 867 阅读

问题:设函数$f$在区间$[a,b]$上处处大于$0$,且对于$L>0$满足Lipschitz条件
\[ |f(x_1)-f(x_2)|\leq L|x_{1}-x_{2}| \]
又已知对于$a\leq c\leq d\leq b$有
\[ \int_{c}^{d}\frac{dx}{f(x)}=\alpha,\qquad \int_{a}^{b}\frac{dx}{f(x)}=\beta \]
证明下列积分不等式:
\[ \int_{a}^{b}f(x)dx\leq \frac{e^{2L\beta}-1}{2L\alpha}\int_{c}^{d}f(x)dx \]

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证明

\[ f(x_{0})=\min_{x\in[a,b]}f(x)=m \]

\[ m\leq f(x)\leq m+L|x-x_{0}| \]
就有
\begin{align*}
&2L\int_{c}^{d}\frac{dx}{f(x)}\cdot\int_{a}^{b}f(x)dx\\
&\leq 2L\int_{c}^{d}\frac{dx}{f(x)}\cdot\int_{a}^{b}[m+L|x-x_{0}|]dx\\
&= 2L\int_{c}^{d}\frac{dx}{f(x)}\cdot\left[\int_{a}^{x_{0}}(m+L(x_{0}-x)dx+\int_{x_{0}}^{b}(m+L(x-x_{0}))dx\right]\\
&= 2L\int_{c}^{d}\frac{dx}{f(x)}\left[m(b-a)+\frac{L}{2}[(x_{0}-a)^2+(b-x_{0})^{2}]\right]\\
&\leq m^{2}\int_{c}^{d}\frac{dx}{f(x)}\left[\frac{2L(b-a)}{m}+\frac{L^2(x_{0}-a)^2+L^2(b-x_{0})^{2}}{m^2}\right]\\
&\leq m(d-c)\left[\frac{2L(b-a)}{m}+\frac{L^2(x_{0}-a)^2+L^2(b-x_{0})^{2}}{m^2}\right]\\
&\leq \int_{c}^{d}f(x)dx\left[\frac{2L(b-a)}{m}+\frac{L^2(x_{0}-a)^2+L^2(b-x_{0})^{2}}{m^2}\right]\\
&=\int_{c}^{d}f(x)dx\left[\frac{2L(b-x_{0})}{m}+\frac{2L(x_{0}-a)}{m}+\frac{L^2(x_{0}-a)^2+L^2(b-x_{0})^{2}}{m^2}\right]\\
&=\int_{c}^{d}f(x)dx\left[\left(\frac{L(x_{0}-a)}{m}+1\right)^{2}+\left(\frac{L(b-x_{0})}{m}+1\right)^{2}-2 \right]\\
&\leq \int_{c}^{d}f(x)dx\left[\left(\frac{L(x_{0}-a)}{m}+1\right)^{2}\left(\frac{L(b-x_{0})}{m}+1\right)^{2}-1 \right]
\end{align*}
兹证明
\[ \left[\left(\frac{L(x_{0}-a)}{m}+1\right)^{2}\left(\frac{L(b-x_{0})}{m}+1\right)^{2}-1 \right]\leq e^{2L\beta}-1 \]
就是
\[ \ln\left(\frac{L(x_{0}-a)}{m}+1\right)+\ln\left(\frac{L(b-x_{0})}{m}+1\right)\leq L\beta \]
这时,只要注意到
\begin{align*}
L\beta&=L\int_{a}^{b}\frac{1}{f(x)}dx\\
&\geq L\int_{a}^{b}\frac{1}{m+L|x-x_{0}|}dx\\
&=L\int_{a}^{x_{0}}\frac{1}{m+L(x_{0}-x)}dx+L\int_{x_{0}}^{b}\frac{1}{m+L(x-x_{0})}dx\\
&=\ln\left(\frac{L(x_{0}-a)}{m}+1\right)+\ln\left(\frac{L(b-x_{0})}{m}+1\right)
\end{align*}
就好.


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