一个涉及Beta函数积分不等式的多种推广

Some Extensions of an Integral Inequality with Beta Function

摘要: 本文对南开大学的一个考研积分不等式进行分析，给出了一种解答，并在此基础上对其进行了多种形式的推广，有助于在教学中引导学生进行数学探讨。

Abstract: We analyse the integral inequality of a postgraduate examination at Nankai University, a method of proof is given, and on the basis several extensions have been made to it, it is helpful to guiding students to carry out mathematics research.

南开大学近年的数学考研题中有这样一个积分不等式证明题：设 $f\left(x\right)$ 在 $\left[0,1\right]$ 上连续可微，且 ${\displaystyle {\int }_0^{1}f\left(x\right)\text{d}x}={\displaystyle {\int }_0^{1}xf\left(x\right)\text{d}x}=1$，试证：

${\displaystyle {\int }_0^1{\left|f^{\prime }\left(x\right)\right|}^{\text{3}}\text{d}x}\ge {\left(\frac{128}{3\pi }\right)}^2$. (1)

这个数学证明题有一定困难，一是不等式的右边是一个具体的数字，而不是一个数学式子，提供的解题思路很少。二是条件与结论中涉及的三个被积函数 $f\left(x\right)$， $xf\left(x\right)$ 及 $f^{\prime }\left(x\right)$ 之间有什么关系需要弄清楚。因此其证明有较大难度。下面给出一种利用重积分的证明方法，并在此基础给出多种形式的推广。

式(1)的证明：设 $x\in \left[0,1\right]$，利用分部积分法，有

${\displaystyle {\int }_0^{x}f\left(t\right)\text{d}t}=xf\left(x\right)-{\displaystyle {\int }_0^{x}t{f}^{\prime }\left(t\right)\text{d}t}$,

两边同时积分，得

${\displaystyle {\int }_0^1{\displaystyle {\int }_0^{x}f\left(t\right)\text{d}t}}\text{d}x={\displaystyle {\int }_0^{1}xf\left(x\right)\text{d}x}-{\displaystyle {\int }_0^1{\displaystyle {\int }_0^{x}t{f}^{\prime }\left(t\right)\text{d}t}}\text{d}x$,

交换积分次序，并根据已知条件，有

${\displaystyle {\int }_0^1\left(1-t\right)f\left(t\right)\text{d}t}=1-{\displaystyle {\int }_0^{1}t\left(1-t\right)f^{\prime }\left(t\right)\text{d}t}$,

又因为由已知条件，有

${\displaystyle {\int }_0^1\left(1-t\right)f\left(t\right)\text{d}t}={\displaystyle {\int }_0^{1}f\left(t\right)\text{d}t}-{\displaystyle {\int }_0^{1}tf\left(t\right)\text{d}t}=1-1=0$,

据此并利用Hölder不等式(见 [1] )，有

$1={\displaystyle {\int }_0^{1}t\left(1-t\right)f^{\prime }\left(t\right)\text{d}t}\le {\left({\displaystyle {\int }_0^1{\left[t\left(1-t\right)\right]}^{\frac{3}{2}}\text{d}t}\right)}^{\frac{2}{3}}{\left({\displaystyle {\int }_0^1{\left|f^{\prime }\left(t\right)\right|}^{\text{3}}\text{d}t}\right)}^{\frac{1}{3}}$,

利用Beta函数与Gamma函数(见 [2] )的定义与性质，有

$\begin{array}{c}{\displaystyle {\int }_0^1{\left[t\left(1-t\right)\right]}^{\frac{3}{2}}\text{d}t}={\displaystyle {\int }_0^1{t}^{\frac{5}{2}-1}{\left(1-t\right)}^{\frac{5}{2}-1}\text{d}t}=B\left(\frac{5}{2},\frac{5}{2}\right)=\frac{\Gamma \left(\frac{5}{2}\right)\Gamma \left(\frac{5}{2}\right)}{\Gamma \left(5\right)}\\ =\frac{1}{4!}{\Gamma }^2\left(\frac{3}{2}+1\right)=\frac{1}{4!}{\left(\frac{3}{2}\right)}^2{\Gamma }^2\left(\frac{3}{2}\right)=\frac{1}{4!}\frac{9}{4}{\Gamma }^2\left(\frac{1}{2}+1\right)\\ =\frac{1}{4!}\frac{9}{4}{\left(\frac{1}{2}\right)}^2{\Gamma }^2\left(\frac{1}{2}\right)=\frac{3}{128}\Gamma \left(\frac{1}{2}\right)\Gamma \left(1-\frac{1}{2}\right)\\ =\frac{3}{128}\frac{\pi }{\sin \pi /2}=\frac{3\pi }{128}\end{array}$,

所以

$1\le {\left(\frac{3\pi }{128}\right)}^2{\displaystyle {\int }_0^1{\left|f^{\prime }\left(x\right)\right|}^{\text{3}}\text{d}x}$,

从而(1)式成立。

下面我们利用此证明方法，给出几种推广。

命题1 设 $f\left(x\right)$ 在 $\left[0,1\right]$ 上连续可微， $n\ge 2$，且 ${\displaystyle {\int }_0^{1}f\left(x\right)\text{d}x}={\displaystyle {\int }_0^{1}xf\left(x\right)\text{d}x}=1$，则有

${\displaystyle {\int }_0^1{\left|f^{\prime }\left(x\right)\right|}^n\text{d}x}\ge {\left[\frac{n}{4\left(n+1\right)\left(3n-1\right)}\right]}^{1-n}{B}^{1-n}\left(\frac{1}{n-1},\frac{1}{n-1}\right)$.

证明：由前述证明，可得

${\displaystyle {\int }_0^{1}x\left(1-x\right)f^{\prime }\left(x\right)\text{d}x}=1$,

利用Hölder不等式，有

$1={\displaystyle {\int }_0^{1}x\left(1-x\right)f^{\prime }\left(x\right)\text{d}x}\le {\left({\displaystyle {\int }_0^1{\left|f^{\prime }\left(x\right)\right|}^n}\text{d}x\right)}^{\frac{1}{n}}{\left({\displaystyle {\int }_0^1{\left[x\left(1-x\right)\right]}^{\frac{n}{n-1}}\text{d}x}\right)}^{\frac{n-1}{n}}$,

所以

$1\le {\displaystyle {\int }_0^1{\left|f^{\prime }\left(x\right)\right|}^n\text{d}x}{\left({\displaystyle {\int }_0^1{\left[x\left(1-x\right)\right]}^{\frac{n}{n-1}}\text{d}x}\right)}^{n-1}$,

又因为

$\begin{array}{c}{\displaystyle {\int }_0^1{\left[x\left(1-x\right)\right]}^{\frac{n}{n-1}}\text{d}x}=B\left(\frac{n}{n-1}+1,\frac{n}{n-1}+1\right)=\frac{{\Gamma }^2\left(\frac{n}{n-1}+1\right)}{\Gamma \left(\frac{2n}{n-1}+2\right)}\\ =\frac{{\Gamma }^2\left(\frac{1}{n-1}+2\right)}{\Gamma \left(\frac{2}{n-1}+4\right)}=\frac{n{\Gamma }^2\left(\frac{1}{n-1}\right)}{4\left(n+1\right)\left(3n-1\right)\Gamma \left(\frac{2}{n-1}\right)}\\ =\frac{n}{4\left(n+1\right)\left(3n-1\right)}B\left(\frac{1}{n-1},\frac{1}{n-1}\right),\end{array}$

于是

${\displaystyle {\int }_0^1{\left|f^{\prime }\left(x\right)\right|}^n\text{d}x}\ge {\left({\displaystyle {\int }_0^1{\left[x\left(1-x\right)\right]}^{\frac{n}{n-1}}\text{d}x}\right)}^{1-n}={\left[\frac{n}{4\left(n+1\right)\left(3n-1\right)}\right]}^{1-n}{B}^{1-n}\left(\frac{1}{n-1},\frac{1}{n-1}\right)$.

注：当 $n=3$ 时，便得到式(1)。

命题2 设 $f\left(x\right)$ 在 $\left[0,1\right]$ 上有 $2n+1$ 阶连续导数， $n\ge 0$， $m\ge 2$， ${\displaystyle {\int }_0^{1}f\left(x\right)\text{d}x}=1$，且

${\displaystyle {\int }_0^1{x}^k{f}^{\left(k-1\right)}\left(x\right)\text{d}x}=k!$, $k=1,2,\cdots ,2\left(n+1\right)$.

则有

${\displaystyle {\int }_0^1{\left|f^{\left(2n+1\right)}\left(x\right)\right|}^m\text{d}x}\ge {\left[\left(2n+1\right)!\right]}^m{B}^{1-m}\left(\frac{m}{m-1}\left(2n+1\right)+1,\frac{m}{m-1}+1\right)$.

证明：利用分部积分法，有

$\begin{array}{c}{\displaystyle {\int }_0^{x}f\left(t\right)\text{d}t}=xf\left(x\right)-{\displaystyle {\int }_0^{x}t{f}^{\prime }\left(t\right)\text{d}t}=xf\left(x\right)-\frac{1}{2}{x}^2{f}^{\prime }\left(x\right)+\frac{1}{2}{\displaystyle {\int }_0^x{t}^2{f}^{″}\left(t\right)\text{d}t}\\ =xf\left(x\right)-\frac{1}{2!}{x}^2{f}^{\prime }\left(x\right)+\frac{1}{3!}{x}^3{f}^{″}\left(x\right)-\frac{1}{3!}{\displaystyle {\int }_0^x{t}^3{f}^{‴}\left(t\right)\text{d}t}=\cdots \\ ={\displaystyle \underset{k=1}{\overset{2n+1}{\sum }}{\left(-1\right)}^{k+1}\frac{x^k}{k!}{f}^{\left(k-1\right)}\left(x\right)}-\frac{1}{\left(2n+1\right)!}{\displaystyle {\int }_0^x{t}^{2n+1}{f}^{\left(2n+1\right)}\left(t\right)\text{d}t},\end{array}$

两边同时积分并利用已知条件，得到

$\begin{array}{l}{\displaystyle {\int }_0^1{\displaystyle {\int }_0^{x}f\left(x\right)\text{d}t\text{d}x}}\\ ={\displaystyle \underset{k=1}{\overset{2n+1}{\sum }}{\left(-1\right)}^{k+1}\frac{1}{k!}}{\displaystyle {\int }_0^1{x}^k{f}^{\left(k-1\right)}\left(x\right)\text{d}x}-\frac{1}{\left(2n+1\right)!}{\displaystyle {\int }_0^1{\displaystyle {\int }_0^x{t}^{2n+1}{f}^{\left(2n+1\right)}\left(t\right)\text{d}t\text{d}x}}\\ =1-\frac{1}{\left(2n+1\right)!}{\displaystyle {\int }_0^1\left(1-t\right)t^{2n+1}{f}^{\left(2n+1\right)}\left(t\right)\text{d}t},\end{array}$

又因为

${\displaystyle {\int }_0^1{\displaystyle {\int }_0^{x}f\left(x\right)\text{d}t\text{d}x}}={\displaystyle {\int }_0^1\left(1-t\right)f\left(t\right)\text{d}t}={\displaystyle {\int }_0^{1}f\left(t\right)\text{d}t}-{\displaystyle {\int }_0^{1}tf\left(t\right)\text{d}t}=1-1=0$,

所以有

$\begin{array}{c}\left(2n+1\right)!={\displaystyle {\int }_0^1\left(1-t\right)t^{2n+1}{f}^{\left(2n+1\right)}\left(t\right)\text{d}t}\\ \le {\left({{\displaystyle {\int }_0^1\left|f^{\left(2n+1\right)}\left(x\right)\right|}}^m\text{d}x\right)}^{\frac{1}{m}}{\left({\displaystyle {\int }_0^1{\left[x^{2n+1}\left(1-x\right)\right]}^{\frac{m}{m-1}}\text{d}x}\right)}^{\frac{m-1}{m}}\\ \le {\left({{\displaystyle {\int }_0^1\left|f^{\left(2n+1\right)}\left(x\right)\right|}}^m\text{d}x\right)}^{\frac{1}{m}}{B}^{\frac{m-1}{m}}\left(\frac{m}{m-1}\left(2n+1\right)+1,\frac{m}{m-1}+1\right),\end{array}$

于是

${\displaystyle {\int }_0^1{\left|f^{\left(2n+1\right)}\left(x\right)\right|}^m\text{d}x}\ge {\left[\left(2n+1\right)!\right]}^m{B}^{1-m}\left(\frac{m}{m-1}\left(2n+1\right)+1,\frac{m}{m-1}+1\right)$.

注：当 $n=0$， $m=3$ 时，则可得到式(1)。

命题3 设 $f\left(x\right)$ 在 $\left[0,1\right]$ 上有 $2n+1$ 阶连续导数， $n\ge 0$，且

${\displaystyle {\int }_0^{1}f\left(x\right)\text{d}x}=1$, ${\displaystyle {\int }_0^1{x}^k{f}^{\left(k-1\right)}\left(x\right)\text{d}x}=k!$, $k=1,2,\cdots ,2\left(n+1\right)$

则有

${\displaystyle {\int }_0^1{\left|f^{\left(2n+1\right)}\left(x\right)\right|}^{2n+3}\text{d}x}\ge {\left[2\left(n+2\right)\right]}^{2\left(n+1\right)}{\left[\left(2n+1\right)!\right]}^{2n+3}$.

证明：仿命题2的证明，可得

$\left(2n+1\right)!={\displaystyle {\int }_0^1\left(1-x\right)x^{2n+1}{f}^{\left(2n+1\right)}\left(x\right)\text{d}x}$,

利用涉及多个函数的Hölder不等式(见 [1] )，有

$\begin{array}{c}{\left[\left(2n+1\right)!\right]}^{2n+3}={\left({\displaystyle {\int }_0^1\left(1-x\right)xx\cdots x{f}^{\left(2n+1\right)}\left(x\right)\text{d}x}\right)}^{2n+3}\\ \le \left({\displaystyle \underset{k=1}{\overset{2n+1}{\sum }}{\displaystyle {\int }_0^1{x}^{2n+3}\text{d}x}}\right){\displaystyle {\int }_0^1{\left(1-x\right)}^{2n+3}\text{d}x}{\displaystyle {\int }_0^1{\left|f^{\left(2n+1\right)}\left(x\right)\right|}^{2n+3}\text{d}x}\\ ={\left[\frac{1}{2\left(n+2\right)}\right]}^{2\left(n+1\right)}{\displaystyle {\int }_0^1{\left|f^{\left(2n+1\right)}\left(x\right)\right|}^{2n+3}\text{d}x},\end{array}$

于是得到

${\displaystyle {\int }_0^1{\left|f^{\left(2n+1\right)}\left(x\right)\right|}^{\text{2}n+\text{3}}\text{d}x}\ge {\left[2\left(n+2\right)\right]}^{2\left(n+1\right)}{\left[\left(2n+1\right)!\right]}^{2n+3}$.