Real Analysis

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

Last update: 19 July 2014

Lecture 25

A metric space is a set $X$ with a function $d:X\times X\to {ℝ}_{\ge 0}$ such that

(a)	if $p\in X$ then $d(p,p)=0,$
(b)	if $p,q\in X$ then $d(p,q)\ne 0,$
(c)	if $p,q\in X$ then $d(p,q)=d(q,p),$
(d)	if $p,q,r\in X$ then $d(p,r)\le d(p,q)+d(q,r)\text{.}$

The point of this lecture is to show:

If $X$ is ${ℝ}^n$ and $$d:{ℝ}^n\times {ℝ}^n⟶{ℝ}_{\ge 0}\text{is given by}d(x,y)=|y-x|$$ then the triangle inequality holds: $$|x+y|\le \left|x\right|+\left|y\right|\text{.}$$ or, if $x=p-q$ and $y=q-r$ then $$|p-r|=|p-q+q-r|<|p-q|+|q-r|$$ so that $$d(p,r)\le d(p,q)+d(q,r)$$ and (d) holds.

${ℝ}^n$ will be our favourite example of a metric space.

The triangle and Schwartz inequalities

The inner product on ${ℝ}^n$ is the function $$\begin{array}{ccc}{ℝ}^n\times {ℝ}^n& ⟶& ℝ\\ (x,y)& ⟼& ⟨x,y⟩\end{array}$$ given by $$⟨x,y⟩=(x_1,\dots ,x_n)\left(\begin{array}{c}{y}_1\\ ⋮\\ y_n\end{array}\right)=x_1{y}_1+\cdots +x_n{y}_n=\sum _{i=1}^n{x}_i{y}_i\text{.}$$

The absolute value on ${ℝ}^n$ is the function $$\begin{array}{ccc}{ℝ}^n& ⟶& {ℝ}_{\ge 0}\\ x& ⟼& \left|x\right|\end{array}\text{given by}\left|x\right|=\sqrt{x_1^2+\cdots +x_n^2}=\sqrt{⟨x,x⟩}\text{.}$$ Pictorially, $\left|x\right|$ is the distance from $x=(x_1,\dots ,x_n)$ to the origin $$\begin{array}{c} (2,1,3) x y z \end{array}$$ $$\begin{array}{ccc}{\displaystyle {ℝ}^3}& =& {\displaystyle \left\{(x,y,z)\hspace{0.17em}\right|\hspace{0.17em}x,y,z\in ℝ\}\text{and}}\\ {\displaystyle {ℝ}^n}& =& {\displaystyle \left\{(x_1,\dots ,x_n)\hspace{0.17em}\right|\hspace{0.17em}{x}_1,x_2,\dots ,x_n\in ℝ\},}\\ {\displaystyle {ℝ}^1}& =& {\displaystyle \left\{x\hspace{0.17em}\right|\hspace{0.17em}x\in ℝ\}=ℝ}\\ {\displaystyle {ℝ}^2}& =& {\displaystyle \left\{(a,b)\hspace{0.17em}\right|\hspace{0.17em}a,b\in ℝ\}\text{can be identified with}}\\ {\displaystyle ℂ}& =& {\displaystyle \{a+bi\hspace{0.17em}|\hspace{0.17em}a,b\in ℝ\}\text{.}}\end{array}$$

Lagrange's identity

If $x=(x_1,\dots ,x_n)\in {ℝ}^n$ and $y=(y_1,y_2,\dots ,y_n)\in {ℝ}^n$ then $$\left(\sum _{i=1}^n{x}_i^2\right)\left(\sum _{i=1}^n{y}_i^2\right)-{\left(\sum _{i=1}^n{x}_i{y}_i\right)}^2=\frac{1}{2}\sum _{i,j}{(x_i{y}_j-x_j{y}_i)}^2\text{.}$$

		Proof.
	$$\begin{array}{ccc}{\displaystyle \frac{1}{2}\sum _{i,j=1}^n{(x_i{y}_j-x_j{y}_i)}^2}& =& {\displaystyle \frac{1}{2}\sum _{i,j=1}^n{x}_i^2{y}_j^2-2x_i{y}_j{x}_j{y}_i+x_j^2{y}_i^2}\\ & =& {\displaystyle \frac{1}{2}\sum _{i,j=1}^n{x}_i^2{y}_j^2+\frac{1}{2}\sum _{i,j=1}^n{x}_j^2{y}_i^2-\sum _{j,i=1}^n{x}_i{y}_i{x}_j{y}_j}\\ & =& {\displaystyle \sum _{i,j=1}^n{x}_i^2{y}_j^2-{\left(\sum _{i=1}^n{x}_i{y}_i\right)}^2}\\ & =& {\displaystyle \left(\sum _{i=1}^n{x}_i^2\right)\left(\sum _{j=1}^n{y}_j^2\right)-{\left(\sum _{i=1}^n{x}_i{y}_i\right)}^2\text{.}}\end{array}$$ $\square$

If $n=2,$ $$\begin{array}{ccc}\multicolumn{3}{c}{{\displaystyle \frac{1}{2}({(x_1{y}_1-x_1{y}_1)}^2+{(x_1{y}_2-x_2{y}_1)}^2+{(x_2{y}_1-x_1{y}_2)}^2+{(x_2{y}_2-x_2{y}_2)}^2)\text{.}}}\\ & =& {\displaystyle \dots }\\ & =& {\displaystyle (x_1^2+x_2^2)(y_1^2+y_2^2)-{(x_1{y}_1+x_2{y}_2)}^2\text{.}}\end{array}$$

(The Schwartz inequality) If $x,y\in {ℝ}^n$ then $$⟨x,y⟩\le \left|x\right|\left|y\right|\text{.}$$

		Proof.
	Lagrange's identity tells us $${\left|x\right|}^2{\left|y\right|}^2-{⟨x,y⟩}^2\ge 0\text{.}$$ So $${\left(\left|x\right|\left|y\right|\right)}^2\ge {⟨x,y⟩}^2\text{.}$$ So $$\left|x\right|\left|y\right|\ge ⟨x,y⟩\text{.}$$ $\square$

(The triangle inequality) Let $x,y\in {ℝ}^n\text{.}$ Then $$|x+y|\le \left|x\right|+\left|y\right|\text{.}$$

		Proof.
	$$\begin{array}{ccc}{\displaystyle ⟨x+y,x+y⟩}& =& {\displaystyle ⟨x,x⟩+⟨x,y⟩+⟨y,x⟩+⟨y,y⟩}\\ & =& {\displaystyle {\left|x\right|}^2+2⟨x,y⟩+{\left|y\right|}^2}\\ & \le & {\displaystyle {\left|x\right|}^2+2\left|x\right|\left|y\right|+{\left|y\right|}^2}\\ & =& {\displaystyle {(\left|x\right|+\left|y\right|)}^2\text{.}}\end{array}$$ So $${|x+y|}^2\le {(\left|x\right|+\left|y\right|)}^2\text{.}$$ So $$|x+y|\le \left|x\right|+\left|y\right|\text{.}$$ $\square$

Note that Lagrange's identity works with $ℝ$ replaced by any field, and the Schwartz and triangle inequalities are valid with $ℝ$ replaced by any ordered field.

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100507Lect25.pdf and was given on 7 May 2010.

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