数学分析笔记——积分不等式

MatrixBi
2019年12月31日

数学分析笔记——积分不等式

1​ Cauchy-Schwarz 不等式

设函数 $f \left( x \right),g \left( x \right)\in L[a,b]$, 证明:
$$
\left( \int_{a}^{b}{f \left( x \right)g \left( x \right)}\mathrm{d}x \right)^2\leq\int_{a}^{b}{f^2 \left( x \right)}\mathrm{d}x\int_{a}^{b}{g^2 \left( x \right)}\mathrm{d}x.
$$

证明: 对任意 $x\in \left( a,b \right), \left( tf \left( x \right)-g \left( x \right) \right)^2\geq0$, 从而
$$
\int_{a}^{b}{ \left( tf \left( x \right)-g \left( x \right) \right)^2}\mathrm{d}x\geq0,
$$
展开得
$$
\left( \int_{a}^{b}{f^2 \left( x \right)}\mathrm{d}x \right)t^2-2 \left( \int_{a}^{b}{f \left( x \right)g \left( x \right)}\mathrm{d}x \right)t+\int_{a}^{b}{g^2 \left( x \right)}\mathrm{d}x\geq0
$$
对 $t\in\mathbb{R}$ 成立. 因此
$$
4 \left( \int_{a}^{b}{f \left( x \right)g \left( x \right)}\mathrm{d}x \right)^2-4\int_{a}^{b}{f^2 \left( x \right)}\mathrm{d}x\int_{a}^{b}{g^2 \left( x \right)}\mathrm{d}x\leq0,
$$
即不等式成立.

2​ Young 不等式

设函数 $f \left( x \right)$在$ \left( 0,c \right)$ 上连续且严格递增, $f \left( 0 \right)=0,a\in \left( 0,c \right),b\in \left( 0,f \left( c \right) \right)$. 证明:
$$
ab\leq\int_{0}^{a}{f \left( x \right)}\mathrm{d}x+\int_{0}^{b}{f^{-1} \left( x \right)}\mathrm{d}x.
$$
证明: 由定积分的几何意义可得.

$2’$ 推论: 若 $u,v\geq0,p,q>0,\frac{1}{p}+\frac{1}{q}=1$, 则
$$
uv\leq\frac{u^p}{p}+\frac{v^q}{q}.
$$

3​ Hölder不等式

设函数 $f \left( x \right),g \left( x \right)\in L[a,b],p,q>0,\frac{1}{p}+\frac{1}{q}=1$. 证明:
$$
\int_{a}^{b}{|f \left( x \right)g \left( x \right)|}\mathrm{d}x\leq \left( \int_{a}^{b}{|f \left( x \right)|^p}\mathrm{d}x \right)^{\frac{1}{p}} \left( \int_{a}^{b}{|g \left( x \right)|^q}\mathrm{d}x \right)^{\frac{1}{q}}.
$$

证明: 不妨设右式大于零. 故利用 $2’$ 可以得到
$$
\int_{a}^{b}{\left|\frac{f \left( x \right)}{ \left( \int_{a}^{b}{|f \left( x \right)|^p}\mathrm{d}x \right)^{\frac{1}{p}}}\frac{g \left( x \right)}{ \left( \int_{a}^{b}{|g \left( x \right)|^q}\mathrm{d}x \right)^{\frac{1}{q}}}\right|}\mathrm{d}x\leq\int_{a}^{b}{ \left( \frac{|f \left( x \right)|^p}{p\int_{a}^{b}{|f \left( x \right)|^p}\mathrm{d}x}+\frac{|g \left( x \right)|^q}{q\int_{a}^{b}{|g \left( x \right)|^q}\mathrm{d}x} \right)}\mathrm{d}x=\frac{1}{p}+\frac{1}{q}=1 ,
$$
由此得结论成立.

4 Minkowski 不等式 $ \left( p>1 \right)$

设函数 $f \left( x \right),g \left( x \right)\in L[a,b],p>1$. 证明:
$$
\left( \int_{a}^{b}{|f \left( x \right)+g \left( x \right)|^p}\mathrm{d}x \right)^{\frac{1}{p}}\leq \left( \int_{a}^{b}{|f \left( x \right)|^p}\mathrm{d}x \right)^{\frac{1}{p}}+ \left( \int_{a}^{b}{|g \left( x \right)|^p}\mathrm{d}x \right)^{\frac{1}{p}}.
$$

证明: 设 $q>1$ 满足 $\frac{1}{p}+\frac{1}{q}=1$.
$$
\begin{equation}
\begin{split}
\left( \int_{a}^{b}{|f \left( x \right)+g \left( x \right)|^p}\mathrm{d}x \right)^{\frac{1}{p}\cdot p}&=\int_{a}^{b}{|f \left( x \right)+g \left( x \right)||f \left( x \right)+g \left( x \right)|^{p-1}}\mathrm{d}x\\
&\leq\int_{a}^{b}{|f \left( x \right)||f \left( x \right)+g \left( x \right)|^{p-1}}\mathrm{d}x+\int_{a}^{b}{|g \left( x \right)||f \left( x \right)+g \left( x \right)|^{p-1}}\mathrm{d}x\\
&\leq \left( \int_{a}^{b}{|f \left( x \right)|^p}\mathrm{d}x \right)^{\frac{1}{p}} \left( \int_{a}^{b}{|f \left( x \right)+g \left( x \right)|^{ \left( p-1 \right)q}}\mathrm{d}x \right)^{\frac{1}{q}}\\
&\quad+ \left( \int_{a}^{b}{|g \left( x \right)|^p}\mathrm{d}x \right)^{\frac{1}{p}} \left( \int_{a}^{b}{|f \left( x \right)+g \left( x \right)|^{ \left( p-1 \right)q}}\mathrm{d}x \right)^{\frac{1}{q}}\\
&= \left( \left( \int_{a}^{b}{|f \left( x \right)|^p}\mathrm{d}x \right)^{\frac{1}{p}}+ \left( \int_{a}^{b}{|g \left( x \right)|^p}\mathrm{d}x \right)^{\frac{1}{p}} \right) \left( \int_{a}^{b}{|f \left( x \right)+g \left( x \right)|^{p}}\mathrm{d}x \right)^{\frac{1}{q}}.\\
\end{split}
\end{equation}
$$
由此得结论成立.

5 Minkowski 不等式 $ \left( 0 < p < 1 \right)$

设非负函数 $f \left( x \right),g \left( x \right)\in L[a,b],0 < p < 1$. 证明: $$ \left( \int_{a}^{b}{ \left( f \left( x \right)+g \left( x \right) \right)^p}\mathrm{d}x \right)^{\frac{1}{p}}\geq \left( \int_{a}^{b}{ \left( f \left( x \right) \right)^p}\mathrm{d}x \right)^{\frac{1}{p}}+ \left( \int_{a}^{b}{ \left( g \left( x \right) \right)^p}\mathrm{d}x \right)^{\frac{1}{p}}. $$ 证明: 不妨设右式二者均大于零. 设 $c= \left( \int_{a}^{b}{ \left( f \left( x \right) \right)^p}\mathrm{d}x \right)^{\frac{1}{p}},d= \left( \int_{a}^{b}{ \left( g \left( x \right) \right)^p}\mathrm{d}x \right)^{\frac{1}{p}}$. 由于 $y=x^p$ 为凹函数. $$ \begin{equation} \begin{split} \left(\int_{a}^{b}{\left(\frac{f(x)+g(x)}{c+d}\right)^p}\mathrm{d}x\right)^{\frac{1}{p}}&=\left(\int_{a}^{b}{\left[\frac{c}{c+d}\frac{f(x)}{c}+\frac{d}{c+d}\frac{g(x)}{d}\right]^p}\mathrm{d}x\right)^{\frac{1}{p}}\\ &\geq\left( \int_{a}^{b}{\left[ \frac{c}{c+d} \frac{(f(x))^p}{\int_{a}^{b}{(f(x))^p}\mathrm{d}x}+ \frac{d}{c+d} \frac{(g(x))^p}{\int_{a}^{b}{(g(x))^p}\mathrm{d}x} \right]} \mathrm{d}x\right) ^{\frac{1}{p}}\\ &=1. \end{split} \end{equation} $$

6​ Chebyshev 不等式

设函数 $f \left( x \right),g \left( x \right)$ 在 $ \left( a,b \right)$ 上单调递增, 证明:
$$
\int_{a}^{b}{f \left( x \right)g \left( x \right)}\mathrm{d}x\geq\frac{1}{b-a}\int_{a}^{b}{f \left( x \right)}\mathrm{d}x\int_{a}^{b}{g \left( x \right)}\mathrm{d}x.
$$

证明: 对任意 $x,y\in \left( a,b \right), \left( f \left( x \right)-f \left( y \right) \right) \left( g \left( x \right)-g \left( y \right) \right)\geq0$, 在 $ \left( a,b \right)$ 上依次分别对 $x,y$ 积分即得结论成立.

 数学分析 笔记