不等式

利用 $e^x$ 下凸, 得到不等式 $a^{1/p}b^{1/p'}\leqslant \frac a{p}+\frac b{p'}$.

取

$$a=\frac{|f(x)|^p}{\Vert f\Vert_p^p},b=\frac{|g(x)|^{p'}}{\Vert g\Vert_{p'}^{p'}}$$

可得

$$\frac{|f(x)g(x)|}{\Vert f\Vert_p\cdot\Vert g\Vert_{p'}}\leqslant \frac 1 p \cdot\frac{|f(x)|^p}{\Vert f\Vert_p^p}+\frac 1{p'}\cdot\frac{|g(x)|^{p'}}{\Vert g\Vert_{p'}^{p'}}$$

对上式两侧积分可得

$$\frac{\Vert fg\Vert_1}{\Vert f\Vert_p\cdot\Vert g\Vert_{p'}}\leqslant 1$$

故不等式成立.

利用 Holder 不等式

$$\begin{aligned} & \int_E |f(x)g(x)|^r\text{d} x=\int_E |f(x)|^r |g(x)|^r\text{d} x\\ &\leqslant \left(\int_E (|f(x)|^r)^{p/r}\text{d} x\right)^{r/p}\left(\int_E(|g(x)|^r)^{r/q}\text{d} x\right)^{r/q} \end{aligned}$$

再对上式开 $r$ 次根号即可得到

$$\Vert fg\Vert_r\leqslant\Vert f\Vert_p\cdot\Vert g\Vert_q.$$

$$\begin{aligned} &\int_E|f(x)+g(x)|^p\text{d} x\\ \leqslant & \int_E|f(x)+g(x)|^{p-1}\cdot|f(x)+g(x)|\text{d} x\\ \leqslant\int_E |f(x)+g(x)|^{p-1}\cdot|f(x)|\text{d} x \\ &+ \int_E |f(x)+g(x)|^{p-1}\cdot |g(x)|\text{d} x \end{aligned}$$

取共轭指标 $p$ 和 $p'=\frac{p}{p-1}$ 对上式两个求和分别使用 Holder 不等式.

$$\int_E |f(x)+g(x)|^{p-1}\cdot|f(x)|\text{d} x\leqslant \Vert f+g\Vert_p^{p-1}+\Vert f\Vert_p$$

对另一部分同样如此, 再将 $\Vert f+g\Vert_p^{p-1}$ 约去即可.