Hölder 不等式

Let $1<p,q<\infty ,\frac{1}{p}+\frac{1}{q}=1$.

Form of Integral
$f\in L^p\left[a,b\right],g\in L^q\left[a,b\right]$, then:

• $$f\left(x\right)g\left(x\right)\in L^1\left[a,b\right]$$
• $${\int }_a^b\left|f\left(x\right)g\left(x\right)\right|dx\le {\left({\int }_a^b{\left|f\left(x\right)\right|}^{p}dx\right)}^{\frac{1}{p}}{\left({\int }_a^b{\left|g\left(x\right)\right|}^{q}dx\right)}^{\frac{1}{q}}$$
  or $${\Vert fg\Vert }_1\le {\Vert f\Vert }_p{\Vert g\Vert }_q$$

Form of Series
$x=\left\{x_k\right\}\in l^p,y=\left\{y_k\right\}\in l^q$, then

• $$z=\left\{x_k{y}_k\right\}\in l^1$$
• $$\sum _{k=1}^{\infty }\left|x_k{y}_k\right|\le {\left(\sum _{k=1}^{\infty }{\left|x_k\right|}^p\right)}^{\frac{1}{p}}{\left(\sum _{k=1}^{\infty }{\left|y_k\right|}^q\right)}^{\frac{1}{q}}$$
  or$${\Vert xy\Vert }_1\le {\Vert x\Vert }_p{\Vert y\Vert }_q$$
  
  

Proof (Use Integral Form of an example)

1. If ${\Vert f\Vert }_p=0\vee {\Vert g\Vert }_q=0$ is true, the conclusion is obvious.
2. Suppose ${\Vert f\Vert }_p>0\&{\Vert g\Vert }_q>0$. Let $\varphi \left(t\right)=t^{\frac{1}{p}}\left(t>0\right)$.
   $${\varphi }^{\prime \prime }\left(t\right)=\frac{1}{p}\left(\frac{1}{p}-1\right)t^{\frac{1}{p}-2}<0,\forall t>0$$
   Hence $\varphi$ is concave and we have
   $$\begin{array}{rl}& \varphi \left(t_1\right)\le \varphi \left(t_2\right)+{\varphi }^{\prime }\left(t_2\right)\left(t_2-t_1\right),\forall t_1,t_2>0\\ & \Rightarrow \varphi \left(t\right)\le \varphi \left(1\right)+{\varphi }^{\prime }\left(1\right)\left(t-1\right)\left(Lett=t_1,t_2=1\right)\\ & \Rightarrow t^{\frac{1}{p}}\le 1+\frac{t-1}{p}=\frac{t}{p}+\frac{1}{q}\left(1\right)\end{array}$$
   Let $u\ge 0,v>0$, then we can set $t=u^p{v}^{-q}>0$.
   $$\begin{array}{rl}& \left(1\right)\Rightarrow {\left(u^p{v}^{-q}\right)}^{\frac{1}{p}}\le \frac{1}{p}\left(u^p{v}^{-q}\right)+\frac{1}{q}\\ & \Rightarrow u{v}^{-\frac{q}{p}}\le \frac{1}{p}{u}^p{v}^{-q}+\frac{1}{q}\\ & \Rightarrow uv\le \frac{1}{p}{u}^p{v}^{-q}\left(v^{\frac{q}{p}+1}\right)+\frac{1}{q}\left(v^{\frac{q}{p}+1}\right)\\ & \Rightarrow uv\le \frac{1}{p}{u}^p{v}^{\frac{q}{p}+1-q}+\frac{1}{q}{v}^{\frac{q}{p}+1}=\frac{u^p}{p}+\frac{v^q}{q}\left(\because \frac{1}{p}+\frac{1}{q}=1\Rightarrow \frac{q}{p}+1=q\right)\end{array}$$
   Let $f_1={{\Vert f\Vert }_p}^{-1}f,g_1={{\Vert g\Vert }_q}^{-1}g$. Then ${\Vert f_1\Vert }_p={\Vert g_1\Vert }_q=1$, and we have
   $$\begin{array}{rl}& {\Vert f_1{g}_1\Vert }_1\\ & ={\Vert {{\Vert f\Vert }_p}^{-1}{{\Vert g\Vert }_q}^{-1}fg\Vert }_1\\ & =\frac{{\Vert fg\Vert }_1}{{\Vert f\Vert }_p{\Vert g\Vert }_q}\left(2\right)\end{array}$$
   Set $u=\left|f_1\right|\&v=\left|g_1\right|$, and we have
   $$\left|f_1\left(x\right)g_1\left(x\right)\right|\le \frac{{\left|f_1\left(x\right)\right|}^p}{p}+\frac{{\left|g_1\left(x\right)\right|}^q}{q}$$
   Therefore, we have
   $$\begin{array}{rl}& {\Vert f_1{g}_1\Vert }_1\\ & ={\left({\int }_a^b{\left|f_1\left(x\right)g_1\left(x\right)\right|}^{1}dx\right)}^{\frac{1}{1}}\\ & \le {\int }_a^b\left[\left(\frac{{\left|f_1\left(x\right)\right|}^p}{p}\right)+\left(\frac{{\left|g_1\left(x\right)\right|}^q}{q}\right)\right]dx\\ & \text{=}\frac{1}{p}{\int }_a^b{\left|f_1\left(x\right)\right|}^{p}dx+\frac{1}{q}{\int }_a^b{\left|g_1\left(x\right)\right|}^{q}dx\\ & =\frac{1}{p}{{\Vert f_1\Vert }_p}^{\frac{1}{p}}+\frac{1}{q}{{\Vert g_1\Vert }_q}^{\frac{1}{q}}\\ & =\frac{1}{p}+\frac{1}{q}=1\left(3\right)\end{array}$$
   $\left(2\right)\&\left(3\right)\Rightarrow {\Vert fg\Vert }_1\le {\Vert f\Vert }_p{\Vert g\Vert }_q$
   
   

Minkowski 不等式

Let $1\le p<\infty$。

Form of Integral
$f,g\in L^p\left[a,b\right]$, then we have

• $$f+g\in L^p\left[a,b\right]$$
• $${\left({\int }_a^b{\left|f\left(x\right)+g\left(x\right)\right|}^{p}dx\right)}^{\frac{1}{p}}\le {\left({\int }_a^b{\left|f\left(x\right)\right|}^{p}dx\right)}^{\frac{1}{p}}+{\left({\int }_a^b{\left|g\left(x\right)\right|}^{p}dx\right)}^{\frac{1}{p}}$$
  or
  $${\Vert f+g\Vert }_p\le {\Vert f\Vert }_p+{\Vert g\Vert }_p$$

Form of Series
$x=\left\{x_k\right\},y=\left\{y_k\right\}\in l^p$, then we have

• $$z=\left\{x_k+y_k\right\}\in l^p$$
• $${\left(\sum _{k=1}^{\infty }{\left|x_k+y_k\right|}^p\right)}^{\frac{1}{p}}\le {\left(\sum _{k=1}^{\infty }{\left|x_k\right|}^p\right)}^{\frac{1}{p}}+{\left(\sum _{k=1}^{\infty }{\left|y_k\right|}^p\right)}^{\frac{1}{p}}$$
  or
  $${\Vert x+y\Vert }_p\le {\Vert x\Vert }_p+{\Vert y\Vert }_p$$
  
  

Proof (Use Form of Integral as an example)
Note that
$$\begin{array}{rl}& {\int }_a^b{\left|f+g\right|}^{p}dx\\ & ={\int }_a^b\left|f+g\right|\cdot {\left|f+g\right|}^{p-1}dx\\ & \le {\int }_a^b\left(\left|f\right|+\left|g\right|\right){\left|f+g\right|}^{p-1}dx\\ & ={\int }_a^b\left|f\right|{\left|f+g\right|}^{p-1}dx+{\int }_a^b\left|g\right|{\left|f+g\right|}^{p-1}dx\end{array}$$
By Hölder’s Inequalty, we have
$$\begin{array}{rl}& {\int }_a^b\left|f\right|{\left|f+g\right|}^{p-1}dx\le {\left({\int }_a^b{\left|f\right|}^{p}dx\right)}^{\frac{1}{p}}{\left({\int }_a^b{\left|f+g\right|}^{\left(p-1\right)q}dx\right)}^{\frac{1}{q}}\\ & {\int }_a^b\left|g\right|{\left|f+g\right|}^{p-1}dx\le {\left({\int }_a^b{\left|g\right|}^{p}dx\right)}^{\frac{1}{p}}{\left({\int }_a^b{\left|f+g\right|}^{\left(p-1\right)q}dx\right)}^{\frac{1}{q}}\end{array}$$
Thus we have
$$\begin{array}{rl}& {\int }_a^b{\left|f+g\right|}^{p}dx\le \left({\Vert f\Vert }_p+{\Vert g\Vert }_p\right){\left({\int }_a^b{\left|f+g\right|}^{\left(p-1\right)q}dx\right)}^{\frac{1}{q}}\\ & =\left({\Vert f\Vert }_p+{\Vert g\Vert }_p\right){\left({\int }_a^b{\left|f+g\right|}^{p}dx\right)}^{\frac{1}{q}}\\ & \left(\because \frac{1}{p}+\frac{1}{q}=1\Rightarrow q+p=pq\Rightarrow q\left(p-1\right)=p\right)\end{array}$$

1. If ${\Vert f+g\Vert }_p={\left({\int }_a^b{\left|f+g\right|}^{p}dx\right)}^{\frac{1}{p}}=0$, then ${\int }_a^b{\left|f+g\right|}^{p}dx=0$, both sides are 0, and we have
   $$\begin{array}{rl}& {\Vert f\Vert }_p+{\Vert g\Vert }_p\\ & ={\left({\int }_a^b{\left|f\left(x\right)\right|}^{p}dx\right)}^{\frac{1}{p}}+{\left({\int }_a^b{\left|g\left(x\right)\right|}^{p}dx\right)}^{\frac{1}{p}}\\ & \ge 0+0\\ & =0\\ & ={\Vert f+g\Vert }_p\end{array}$$

2. Otherwise, one divides both sides by ${\left({\int }_a^b{\left|f+g\right|}^{p}dx\right)}^{\frac{1}{q}}$, we have
   $$\begin{array}{rl}& {\left({\int }_a^b{\left|f+g\right|}^{p}dx\right)}^{1-\frac{1}{q}}\le {\Vert f\Vert }_p+{\Vert g\Vert }_p\\ & \Rightarrow {\left({\int }_a^b{\left|f+g\right|}^{p}dx\right)}^{\frac{1}{p}}={\Vert f+g\Vert }_p\le {\Vert f\Vert }_p+{\Vert g\Vert }_p\end{array}$$