闵可夫斯基不等式

向量范数形式

$$\Vert \mathbf{x}{\Vert }_p=(\sum _{i=1}^n{\left|x_i\right|}^p{)}^{\frac{1}{p}}$$

p>=1

当$p\ge 1$,则
$$\Vert \mathbf{x}+\mathbf{y}{\Vert }_p\le \Vert \mathbf{x}{\Vert }_p+\Vert \mathbf{y}{\Vert }_p$$
其实等价于证明p范数是一个向量范数

证明:
$p=1$时,由三角不等式，显然成立
$p>1$
设$q=\frac{p}{p-1}>1$
$$\begin{array}{rl}& \sum _{i=1}^n{\left|x_i+y_i\right|}^p\\ & =\sum _{i=1}^n\left|x_i+y_i\right|{\left|x_i+y_i\right|}^{p-1}\\ & \le \sum _{i=1}^n(\left|x_i\right|+\left|y_i\right|){\left|x_i+y_i\right|}^{p-1}\\ & =\sum _{i=1}^n\left|x_i\right|{\left|x_i+y_i\right|}^{p-1}+\sum _{i=1}^n\left|y_i\right|{\left|x_i+y_i\right|}^{p-1}\\ & \le (\sum _{i=1}^n{\left|x_i\right|}^p{)}^{\frac{1}{p}}(\sum _{i=1}^n{\left|x_i+y_i\right|}^{(p-1)q}{)}^{\frac{1}{q}}\\ & +(\sum _{i=1}^n{\left|y_i\right|}^p{)}^{\frac{1}{p}}(\sum _{i=1}^n{\left|x_i+y_i\right|}^{(p-1)q}{)}^{\frac{1}{q}}\\ & =((\sum _{i=1}^n{\left|x_i\right|}^p{)}^{\frac{1}{p}}+(\sum _{i=1}^n{\left|y_i\right|}^p{)}^{\frac{1}{p}})(\sum _{i=1}^n{\left|x_i+y_i\right|}^p{)}^{\frac{1}{q}}\\ & =(\Vert \mathbf{x}{\Vert }_p+\Vert \mathbf{y}{\Vert }_p)(\sum _{i=1}^n{\left|x_i+y_i\right|}^p{)}^{\frac{1}{q}}\end{array}$$
其中最后一个小于等于号用的Holder不等式

观察首尾
$$\Rightarrow \Vert \mathbf{x}+\mathbf{y}{\Vert }_p\le \Vert \mathbf{x}{\Vert }_p+\Vert \mathbf{y}{\Vert }_p$$

0

当$p\ge 1,x_i,y_i>0$,则
$$\Vert \mathbf{x}+\mathbf{y}{\Vert }_p\ge \Vert \mathbf{x}{\Vert }_p+\Vert \mathbf{y}{\Vert }_p$$
$f(x)=x^p$
$f^{\prime \prime }(x)=p(p-1)x^{p-2}<0$
由Jensen不等式
$$\begin{array}{rl}\sum _{i=1}^n(x_i+y_i{)}^p& =\sum _{i=1}^n(t\frac{x_i}{t}+(1-t)\frac{y_i}{1-t}{)}^p\\ & \ge t\sum _{i=1}^n((\frac{x_i}{t}{)}^p+(1-t)\sum _{i=1}^n(\frac{y_i}{1-t}{)}^p)\\ & =t^{1-p}\sum _{i=1}^n{x}_i^p+(1-t{)}^{1-p}\sum _{i=1}^n{y}_i^p\\ & =t^{1-p}\Vert \mathbf{x}{\Vert }_p^p+(1-t{)}^{1-p}\Vert \mathbf{y}{\Vert }_p^p\\ & =(\frac{\Vert \mathbf{x}{\Vert }_p}{\Vert \mathbf{x}{\Vert }_p+\Vert \mathbf{y}{\Vert }_p}{)}^{1-p}\Vert \mathbf{x}{\Vert }_p^p\\ & +(1-\frac{\Vert \mathbf{x}{\Vert }_p}{\Vert \mathbf{x}{\Vert }_p+\Vert \mathbf{y}{\Vert }_p}{)}^{1-p}\Vert \mathbf{y}{\Vert }_p^p\\ & =(\Vert \mathbf{x}{\Vert }_p+\Vert \mathbf{y}{\Vert }_p{)}^p\end{array}$$
$$\Rightarrow \Vert \mathbf{x}+\mathbf{y}{\Vert }_p\ge \Vert \mathbf{x}{\Vert }_p+\Vert \mathbf{y}{\Vert }_p$$