dual norm conjugate

Young’s inequality

$$ab\le \frac{a^p}{p}+\frac{b^q}{q}$$
Where $a\ge 0,b\ge 0,p>1,q>1,\frac{1}{p}+\frac{1}{q}=1$
if and only if$a^p=b^{}$Wait when${}^q$

Proof:
when $a=0$ or$b=$It is obviously established at$0$

当$a>0,b>0Constant$
Jensen's Coefficientln
$$\begin{array}{r}\ln \left(\frac{a^p}{p}+\frac{b^q}{q}\right)\ge \frac{1}{p}\ln a^p+\frac{1}{q}\ln b^q=\ln ab\end{array}$$
if and only if $a^p=b^{}$Wait when${}^q$

Hölder's inequality

$$\sum _{i=1}^n\left|x_i{y}_i\right|\le \Vert \mathbf{x}{\Vert }_p\Vert \mathbf{y}{\Vert }_q$$
Where $p\ge 1,q\ge 1,\frac{1}{p}+\frac{1}{q}=1$
$\Vert x{\Vert }_p={\left({\sum }_{i=1}^n{|x_i|}^p\right)}^{\frac{1}{p}}$

If and only if $\mathbf{x}=0$或$\mathbf{y}=0$,或$\exists c1,c2>0$，使得$c_1{\left|x_i\right|}^p=c_2{|y_i|}^q$

Proof:
when $\mathbf{x}=0$或$y=$It is obviously established at$0$

当$x\ne 0,y\ne 0$ o'clock

When $p=1$时
${\sum }_{i=1}^n\left|x_i{y}_i\right|\le \left({\max }_i\left|y_i\right|\right){\sum }_{i=1}^n|x_i|=\Vert \mathbf{x}{\Vert }_1\Vert \mathbf{y}{\Vert }_{\infty }$
In the same way $q=1$

When $p>1$ hour

设$A_i=\frac{|x_i|}{\Vert x{\Vert }_p},B_i=\frac{|y_i|}{\Vert \mathbf{y}{\Vert }_q}$

由Young不等式
$$A_i{B}_i=\frac{|x_i{y}_i|}{\Vert \mathbf{x}{\Vert }_p\Vert y{\Vert }_q}\le \frac{1}{p}\frac{{|x_i|}^p}{\Vert x{\Vert }_p^p}+\frac{1}{q}\frac{{|y_i|}^q}{\Vert y{\Vert }_q^q}$$

Sum of both sides
$$\begin{array}{rl}\sum _{i=1}^n\frac{\left|x_i{y}_i\right|}{\Vert \mathbf{x}{\Vert }_p\Vert \mathbf{y}{\Vert }_q}& \le \sum _{i=1}^n\frac{1}{p}\frac{{|x_i|}^p}{\Vert x{\Vert }_p^p}+\frac{1}{q}\frac{{|y_i|}^q}{\Vert y{\Vert }_q^q}\\ \sum _{i=1}^n\frac{|x_i{y}_i|}{\Vert x{\Vert }_p\Vert y{\Vert }_q}& \le \frac{1}{p}\frac{{\sum }_{i=1}^n{|x_i|}^p}{\Vert x{\Vert }_p^p}+\frac{1}{q}\frac{{\sum }_{i=1}^n{|y_i|}^q}{\Vert y{\Vert }_q^q}\\ \sum _{i=1}^n\frac{|x_i{y}_i|}{\Vert x{\Vert }_p\Vert y{\Vert }_q}& \le \frac{1}{p}\frac{\Vert x{\Vert }_p^p}{\Vert x{\Vert }_p^p}+\frac{1}{q}\frac{\Vert y{\Vert }_q^q}{\Vert y{\Vert }_q^q}\\ \sum _{i=1}^n\frac{|x_i{y}_i|}{\Vert x{\Vert }_p\Vert y{\Vert }_q}& \le \frac{1}{p}+\frac{1}{q}\\ \sum _{i=1}^n|x_i{y}_i|& \le \Vert x{\Vert }_p\Vert y{\Vert }_q\end{array}$$

$$\begin{array}{rl}{c}_1{|x_i|}^p& =c_2{|y_i|}^q\\ c_1\sum _{i=1}^n{|x_i|}^p& =c_2\sum _{i=1}^n{|y_i|}^q\\ \frac{c_1}{c_2}& =\frac{\Vert y{\Vert }_q^q}{\Vert x{\Vert }_p^p}\end{array}$$

$$\begin{array}{rl}{c}_1{\left|x_i\right|}^p& =c_2{|y_i|}^q\\ |y_i|& =\sqrt[q]{\frac{c_1}{c_2}{|x_i|}^p}\\ |y_i|& =\frac{\Vert \mathbf{y}{\Vert }_q}{\Vert \mathbf{x}{\Vert }_p^{p-1}}{|x_i|}^{p-1}\\ |x_i{y}_i|& =\frac{\Vert y{\Vert }_q}{\Vert x{\Vert }_p^{p-1}}{|x_i|}^p\\ \sum _{i=1}^n|x_i{y}_i|& =\sum _{i=1}^n\frac{\Vert y{\Vert }_q}{\Vert x{\Vert }_p^{p-1}}{|x_i|}^p\\ \sum _{i=1}^n|x_i{y}_i|& =\Vert x{\Vert }_p\Vert y{\Vert }_q\end{array}$$

dual norm of p norm

Let $\Vert \cdot \Vert$ is defined in${\mathbb{R}}^{}$Norm of${}^n$
Dual norm$\Vert \cdot {\Vert }_{\ast }$定义为
$$\Vert \mathbf{z}{\Vert }_{\ast }=\underset{\Vert x\Vert \le 1}{\sup }{\mathbf{z}}^T\mathbf{x}$$

It can also be derived that ${\mathbf{x}}^T\mathbf{z}\le \Vert \mathbf{x}\Vert \Vert \mathbf{z}{\Vert }_{\ast }$

p norm

$$\Vert \mathbf{z}{\Vert }_q=\underset{\Vert x{\Vert }_p\le 1}{\sup }{z}^{T}x$$
where$p\ge 1,q\ge 1,\frac{1}{p}+\frac{1}{q}=1$

prove:

当$z=0$ , it is obviously established

当$z\ne$Let$0$
hold a function in
$$\begin{array}{r}{\mathbf{z}}^T\mathbf{x}\le \sum _{i=1}^n\left|z_i\right||x_i|\le \Vert \mathbf{x}{\Vert }_p\Vert \mathbf{z}{\Vert }_q\le \Vert z{\Vert }_q\end{array}$$

Let's verify the equality. Let
xi $x_i=\frac{\mathrm{sign}\left(z_i\right){|z_i|}^{q-1}}{\Vert z{\Vert }_q^{\frac{q}{p}}}$
$$\begin{array}{rl}\Vert \mathbf{x}{\Vert }_p^p& =\sum _{i=1}^n{\left|\frac{\mathrm{sign}\left(z_i\right){\left|z_i\right|}^{q-1}}{\Vert \mathbf{z}{\Vert }_q^{\frac{q}{p}}}\right|}^p\\ & =\sum _{i=1}^n\frac{{|z_i|}^{pq-}}{\Vert z{\Vert }_q^q}\\ & =\sum _{i=1}^n\frac{{\left|z_i\right|}^q}{\Vert \mathbf{z}{\Vert }_q^q}\\ & =\frac{\Vert z{\Vert }_q^q}{\Vert z{\Vert }_q^q}\\ & =1\end{array}$$
So
$\Vert \mathbf{x}{\Vert }_p\le 1$
因此
$$\begin{array}{rl}{\mathbf{z}}^T\mathbf{x}& =\sum _{i=1}^n\frac{\mathrm{sign}\left(z_i\right){\left|z_i\right|}^{q-1}}{\Vert \mathbf{z}{\Vert }_q^{\frac{q}{p}}}{z}_i\\ & =\sum _{i=1}^n\frac{{|z_i|}^q}{\Vert z{\Vert }_q^{\frac{q}{p}}}\\ & =\frac{\Vert z{\Vert }_q^q}{\Vert z{\Vert }_q^{\frac{q}{p}}}\\ & =\Vert z{\Vert }_q\end{array}$$

conjugate

$f:{\mathbb{R}}^n\to R$
equation
$$f^{\ast }\left(\mathbf{y}\right)=\underset{\mathbf{x}}{\max }{\mathbf{y}}^T\mathbf{x}-f\left(\mathbf{x}\right)$$

norm

$f\left(x\right)=\Vert x\Vert$
$$f^{\ast }\left(y\right)=\{\begin{array}{ll}0,& \Vert \mathbf{y}{\Vert }_{\ast }\le 1\\ +\infty ,& \Vert y{\Vert }_{\ast }>1\end{array}$$

Among them $\Vert \cdot {\Vert }_{\ast }$is the dual norm

Proof:
When $\Vert \mathbf{y}{\Vert }_{\ast }\le 1$ hour

$${\mathbf{y}}^T\mathbf{x}\le \Vert \mathbf{x}\Vert \Vert \mathbf{y}{\Vert }_{\ast }\le \Vert \mathbf{x}\Vert$$
取$\mathbf{x}=0$ ,the function
$$f^{\ast }\left(\mathbf{y}\right)=y^{T}x-\Vert \mathbf{x}\Vert =0$$

当$\Vert y{\Vert }_{\ast }>1$ is
defined by the dual norm
$\Vert \mathbf{y}{\Vert }_{\ast }={\max }_{\Vert \mathbf{x}\Vert \le 1}{\mathbf{x}}^T\mathbf{y}>1$ means that x \mathbf{x}
exists$x$ ,let$\Vert \mathbf{x}\Vert \le 1,x^{T}y>1$

$$\begin{array}{rl}{f}^{\ast }\left(\mathbf{y}\right)& =\underset{\mathbf{x}}{\max }{\mathbf{y}}^T\mathbf{x}-\Vert \mathbf{x}\Vert \\ & \ge y^T\left(t\mathbf{x}\right)-t\Vert \mathbf{x}\Vert \\ & =t\left({\mathbf{y}}^{T}x-\Vert \mathbf{x}\Vert \right)\end{array}$$
(The second line is because $x$ is arbitrary, so substitute$tx$ will also be established)

t $t\to \infty$时,$f^{\ast }\left(\mathbf{y}\right)\to \infty$