定积分的性质证明不等式：Holder不等式，Schwarz不等式，Minkowski不等式

• 定积分的性质

(1) 叙述并证明Holder不等式；

(2) 叙述并证明Schwarz不等式；

(3) 叙述并证明Minkowski不等式.

解 (1)

$\text { (Hölder 不等式) 设 } f(x), g(x) \text { 在 }[a, b] \text { 上连续, } p, q \text { 为满足 } \frac{1}{p}+\frac{1}{q}=1 \text { 的 }$

正数, 证明

$\int_{a}^{b}|f(x) g(x)| \mathrm{d} x \leqslant\left(\int_{a}^{b}|f(x)|^{p} \mathrm{~d} x\right) ^{\frac{1}{p}}\left(\int_{a}^{b}|g(x)|^{q} \mathrm{~d} x\right)^{\frac{1}{q}} \text {. }$

证 a. 当 $f(x) \equiv 0$ 或 $g(x) \equiv 0$ 时,上式显然成立.
b. 否则的话, 令

$\varphi(x)=\dfrac{|f(x)|}{\left(\int_{a}^{b}|f(x)|^{p} \mathrm{~d} x\right)^{\frac{1}{p}}}, \psi(x)=\dfrac{|g(x)|}{\left(\int_{a}^{b}|g(x)|^{q} \mathrm{~d} x\right)^{\frac{1}{q}}}, x \in[a, b]$

( $\int_{a}^{b}|f(x)|^{\prime} \mathrm{d} x$ 和 $\int_{a}^{b}|g(x)|^{\prime} \mathrm{d} x$ 均大于零), 得到

$\varphi(x) \psi(x) \leqslant \frac{1}{p} \varphi(x)^{p}+\frac{1}{q} \psi(x)^{q},$

即

$\frac{|f(x) g(x)|}{\left(\int_{a}^{b}|f(x)|^{p} \mathrm{~d} x\right)^{\frac{1}{p}}\left(\int_{a}^{b}|g(x)|^{q} \mathrm{~d} x\right)^{\frac{1}{q}}} \leqslant\frac{|f(x)|^{p}}{p\int_{a}^{b}|f(x)|^{p}\mathrm{d}x}+\frac{|g(x)|^{q}}{q\int_{a}^{b}|g(x)|^{q}\mathrm{d}x},x\in[a,b].$

对上式两边在 $[a, b]$ 上求积分, 利用定积分的性质得

$\begin{aligned} & \frac{1}{\left(\int_{a}^{b}|f(x)|^{p} \mathrm{~d} x\right)^{\frac{1}{p}}\left(\int_{a}^{b}|g(x)|^{q} \mathrm{~d} x\right)^{\frac{1}{q}}} \int_{a}^{b}|f(x) g(x)| \mathrm{d} x \\ \leqslant & \frac{\int_{a}^{b}|f(x)|^{p} \mathrm{~d} x}{p \int_{a}^{b}|f(x)|^{p} \mathrm{~d} x}+\frac{\int_{a}^{b}|g(x)|^{q} \mathrm{~d} x}{q \int_{a}^{b}|f(x)|^{q} \mathrm{~d} x}=\frac{1}{p}+\frac{1}{q}=1 . \end{aligned}$

在不等式两边同乘以 $\left(\int_{a}^{b}|f(x)|^{p} \mathrm{d} x\right)^{\frac{1}{p}}\left(\int_{a}^{b}|g(x)|^{q} \mathrm{~d} x\right)^{\frac{1}{q}}$, 即得

$\int_{a}^{b}|f(x) g(x)| \mathrm{d} x \leqslant\left(\int_{a}^{b}|f(x)|^{p} \mathrm{~d} x\right)^{\frac{1}{p}}\left(\int_{a}^{b}|g(x)|^{q} \mathrm{~d} x\right)^{\frac{1}{q}} .$

(2) (Schwarz 不等式) $\left[\int_{a}^{b} f(x) g(x) \mathrm{d} x\right]^{2} \leqslant \int_{a}^{b} f^{2}(x) \mathrm{d} x \cdot \int_{a}^{b} g^{2}(x) \mathrm{d} x$ ；

证 由于对任意的 $t$, 积分 $\int_{a}^{b}[t f(x)+g(x)]^{2} \mathrm{~d} x \geqslant 0$, 即

$t^{2} \int_{a}^{b} f^{2}(x) \mathrm{d} x+2 t \int_{a}^{b} f(x) g(x) \mathrm{d} x+\int_{a}^{b} g^{2}(x) \mathrm{d} x \geqslant 0,$

所以其判别式恒为非正的, 也就是成立

$\left[\int_{a}^{b} f(x) g(x) \mathrm{d} x\right]^{2} \leqslant \int_{a}^{b} f^{2}(x) \mathrm{d} x \cdot \int_{a}^{b} g^{2}(x) \mathrm{d} x .$

(3) (Minkowski 不等式)

$\left\{\int_{a}^{b}[f(x)+g(x)]^{2} \mathrm{~d} x\right\}^{\frac{1}{2}} \leqslant\left\{\int_{a}^{b} f^{2}(x) \mathrm{d} x\right\}^{\frac{1}{2}}+\left\{\int_{a}^{b} g^{2}(x) \mathrm{d} x\right\}^{\frac{1}{2}} .$

证

$\begin{aligned} &\text { 由 } \int_{a}^{b} f(x) g(x) \mathrm{d} x \leqslant\left\{\int_{a}^{b} f^{2}(x) \mathrm{d} x\right\}^{\frac{1}{2}}\left\{\int_{a}^{b} g^{2}(x) \mathrm{d} x\right\}^{\frac{1}{2}} \text {, 得到 } \\ &\int_{a}^{b} f^{2}(x) \mathrm{d} x+2 \int_{a}^{b} f(x) g(x) \mathrm{d} x+\int_{a}^{b} g^{2}(x) \mathrm{d} x \\ &\leqslant \int_{a}^{b} f^{2}(x) \mathrm{d} x+2\left\{\int_{a}^{b} f^{2}(x) \mathrm{d} x\right\}^{\frac{1}{2}}\left\{\int_{a}^{b} g^{2}(x) \mathrm{d} x\right\}^{\frac{1}{2}}+\int_{a}^{b} g^{2}(x) \mathrm{d} x, \end{aligned}$

即

$\int_{a}^{b}[f(x)+g(x)]^{2} \mathrm{~d} x \leqslant\left[\left\{\int_{a}^{b} f^{2}(x) \mathrm{d} x\right\}^{\frac{1}{2}}+\left\{\int_{a}^{b} g^{2}(x) \mathrm{d} x\right\}^{\frac{1}{2}}\right]^{2},$

两边开平方,即得到

$\left\{\int_{a}^{b}[f(x)+g(x)]^{2} \mathrm{~d} x\right\}^{\frac{1}{2}} \leqslant\left\{\int_{a}^{b} f^{2}(x) \mathrm{d} x\right\}^{\frac{1}{2}}+\left\{\int_{a}^{b} g^{2}(x) \mathrm{d} x\right\}^{\frac{1}{2}} \text {. }$