Metric and Hilbert Spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 4 November 2014

Lecture 3 and 4

Hölder, Minkowski, Cauchy-Schwarz and triangle inequalities

Let $q\in {ℝ}_{\ge 1}\text{.}$ Let $p\in {ℝ}_{>1}\cup \left\{\infty \right\}$ be given by $$\frac{1}{p}+\frac{1}{q}=1\text{.}$$ The functions $h:{ℝ}_{\ge 0}\to {ℝ}_{\ge 0}$ given by $$h\left(x\right)=x^q\begin{array}{c} 1 x 1 y y=x y=x2 y=x3 y=x4 \end{array}$$ are increasing with $h\left(1\right)=1\text{.}$

The functions $h:{ℝ}_{\ge 0}\to {ℝ}_{\ge 0}$ given by $$h\left(x\right)=x^{\frac{1}{q}}\begin{array}{c} 1 x 1 y y=x y=x12 y=x13 y=x14 \end{array}$$ are increasing with $h\left(1\right)=1\text{.}$

The functions $h:{ℝ}_{>0}\to {ℝ}_{\ge 0}$ given by $$h\left(x\right)=x^{-\frac{1}{q}}\begin{array}{c} 1 x 1 y y=1x y=x-12 y=x-13 \end{array}$$ are decreasing with $h\left(1\right)=1\text{.}$

Thus the functions $g:{ℝ}_{\ge 0}\to ℝ$ given by $g\left(x\right)=x^{-\frac{1}{q}}-1$ are decreasing with $g\left(1\right)=0\text{.}$ So $$x^{-\frac{1}{q}}-1\le 0\text{for}x\in {ℝ}_{\ge 1}\text{.}$$ If $f:{ℝ}_{>0}\to ℝ$ is given by $$f\left(x\right)=x^{\frac{1}{p}}-\frac{1}{p}x$$ then $\frac{df}{dx}=\frac{1}{p}{x}^{\frac{1}{p}-1}-\frac{1}{p}=\frac{1}{p}(x^{-\frac{1}{q}}-1)$ and so $f$ is decreasing for $x\in {ℝ}_{>1}$ and $f\left(1\right)=1-\frac{1}{p}=\frac{1}{q}\text{.}$ So $$x^{\frac{1}{p}}-\frac{1}{p}x\le \frac{1}{q}\text{for}x\in {ℝ}_{\ge 1}\text{.}$$

Let $a,b\in {ℝ}_{>0}$ with $a\ge b$ and let $x=\frac{a}{b}\text{.}$ Then $$\frac{1}{q}\ge {\left(\frac{a}{b}\right)}^{\frac{1}{p}}-\frac{1}{p}\left(\frac{a}{b}\right)=\frac{1}{b}(a^{\frac{1}{p}}{b}^{-\frac{1}{p}+1}-\frac{1}{p}a)=\frac{1}{b}(a^{\frac{1}{p}}{b}^{\frac{1}{q}}-\frac{1}{p}a)\text{.}$$ So $$\frac{1}{p}a+\frac{1}{q}b\ge a^{\frac{1}{p}}{b}^{\frac{1}{q}}\text{for}a,b\in {ℝ}_{>0}\text{.}$$

Let $x=(x_1,\dots ,x_n)\in {ℝ}^n$ and $y=(y_1,\dots ,y_n)\in {ℝ}^n\text{.}$ Then $$\frac{\left|x_i{y}_i\right|}{{\Vert x\Vert }_p{\Vert y\Vert }_q}\le \frac{1}{p}{\left(\frac{\left|x_i\right|}{{\Vert x\Vert }_p}\right)}^p+\frac{1}{q}{\left(\frac{\left|y_i\right|}{{\Vert y\Vert }_q}\right)}^q\text{.}$$ So $$\begin{array}{ccc}{\displaystyle \sum _{i=1}^n\frac{\left|x_i{y}_i\right|}{{\Vert x\Vert }_p{\Vert y\Vert }_q}}& \le & {\displaystyle \sum _{i=1}^n(\le \frac{1}{p}{\left(\frac{\left|x_i\right|}{{\Vert x\Vert }_p}\right)}^p+\frac{1}{q}{\left(\frac{\left|y_i\right|}{{\Vert y\Vert }_q}\right)}^q)}\\ & =& {\displaystyle \frac{1}{p}+\frac{1}{q}=1\text{.}}\end{array}$$ So $$\begin{array}{ccc}{\displaystyle \sum _{i=1}^n\left|x_i{y}_i\right|}& \le & {\displaystyle {\Vert x\Vert }_p{\Vert y\Vert }_q\text{.}}\end{array}$$ So $$\left|\sum _{i=1}^n{x}_i{y}_i\right|\le \sum _{i=1}^n\left|x_i{y}_i\right|\le {\Vert x\Vert }_p{\Vert y\Vert }_q\text{.}$$

Using $$|x_i+y_i|\le \left|x_i\right|+\left|y_i\right|\text{and}p-1=p(1-\frac{1}{p})=p\frac{1}{q}=\frac{p}{q}$$ and $$\begin{array}{ccc}{\displaystyle {\Vert {|x_1+y_1|}^{\frac{p}{q}},\dots ,{|x_n+y_n|}^{\frac{p}{q}}\Vert }_q}& =& {\displaystyle {\left(\sum _{i=1}^n{\left({(x_i+y_i)}^{\frac{p}{q}}\right)}^q\right)}^{\frac{1}{q}}}\\ & =& {\displaystyle {\left(\sum _{i=1}^n{|x_i+y_i|}^p\right)}^{\frac{1}{p}·\frac{p}{q}}}\\ & =& {\displaystyle {\left({\Vert x+y\Vert }_p\right)}^{\frac{p}{q}},}\end{array}$$ gives $$\begin{array}{ccc}{\displaystyle {|x+y|}_p^p}& =& {\displaystyle \sum _{i=1}^n{|x_i+y_i|}^p=\sum _{i=1}^n|x_i+y_i|{|x_i+y_i|}^{p-1}}\\ & \le & {\displaystyle \sum _{i=1}^n(\left|x_i\right|+\left|y_i\right|){|x_i+y_i|}^{\frac{p}{q}}}\\ & =& {\displaystyle \sum _{i=1}^n\left|x_i\right|{|x_i+y_i|}^{\frac{p}{q}}+\sum _{i=1}^n\left|y_i\right|{|x_i+y_i|}^{\frac{p}{q}}}\\ & \le & {\displaystyle {\Vert x\Vert }_p{\Vert ({|x_1+y_1|}^{\frac{p}{q}},\dots ,{|x_n+y_n|}^{\frac{p}{q}})\Vert }_q+{\Vert y\Vert }_p{\Vert ({|x_1+y_1|}^{\frac{p}{q}},\dots ,{|x_n+y_n|}^{\frac{p}{q}})\Vert }_q}\\ & =& {\displaystyle {\Vert x\Vert }_p{\Vert x+y\Vert }_p^{\frac{p}{q}}+{\Vert y\Vert }_p{\Vert x+y\Vert }_p^{\frac{p}{q}}}\\ & =& {\displaystyle ({\Vert x\Vert }_p+{\Vert y\Vert }_p){\Vert x+y\Vert }_p^{p-1}\text{.}}\end{array}$$ Dividing both sides by ${\Vert x+y\Vert }_p^{p-1},$ then $${\Vert x+y\Vert }_p\le {\Vert x\Vert }_p+{\Vert y\Vert }_p\text{.}$$

Let $x=(x_1,\dots ,x_n)\in {ℝ}^n\text{.}$ For $p\in {ℝ}_{\ge 1}$ define $${\Vert x\Vert }_p={({\left|x_1\right|}^p+\cdots +{\left|x_n\right|}^p)}^{\frac{1}{p}}\text{.}$$ For $x=(x_1,\dots ,x_n)$ in ${ℝ}^n$ define $$\left|x\right|={\left|x\right|}_2={({\left|x_1\right|}^2+\cdots +{\left|x_n\right|}^2)}^{\frac{1}{2}}=\sqrt{x_1^2+\cdots +x_n^2}$$ and, for $x=(x_1,\dots ,x_n)$ and $y=(y_1,\dots ,y_n)$ in ${ℝ}^n$ define $$⟨x,y⟩=x_1{y}_1+\cdots +x_n{y}_n\text{.}$$

Let $x,y\in {ℝ}^n$ with $x=(x_1,\dots ,x_n)$ and $y=(y_1,\dots ,y_n)\text{.}$ If $p\in {ℝ}_{>1}$ and $q$ is given by $\frac{1}{p}+\frac{1}{q}=1$ then $$\left|\sum _{i=1}^n{x}_i{y}_i\right|\le {\Vert x\Vert }_p{\Vert y\Vert }_q\text{and}{\Vert x+y\Vert }_p\le {\Vert x\Vert }_p+{\Vert y\Vert }_p\text{.}$$

Let $,xy\in {ℝ}^n$ with $x=(x_1,\dots ,x_n)$ and $y=(y_1,\dots ,y_n)\text{.}$ Then $$\left|⟨x,y⟩\right|=\left|\sum _{i=1}^n{x}_i{y}_i\right|\le \left|x\right|\left|y\right|\text{and}|x+y|\le \left|x\right|+\left|y\right|\text{.}$$

Special case: Let $x,y\in ℝ\text{.}$ Then $$\left|xy\right|=\left|x\right|\left|y\right|\text{and}|x+y|\le \left|x\right|+\left|y\right|\text{.}$$

Notes and References

These are a typed copy of Lecture 3 and 4 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on July 31 and August 1, 2014.

page history