Cauchy-Schwarz and triangle inequalities

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

Last updates: 03 January 2012

The space ${ℝ}^n$

Define $${ℝ}^n=\left\{\right(x_1,x_2,\dots ,x_n\left)\right|x_1,x_2,\dots ,x_n\in ℝ\}$$ so that $${ℝ}^1=ℝ,{ℝ}^2=\left\{\right(x,y\left)\right|x,y\in ℝ\},{ℝ}^3=\{(x,y,z)|x,y,z\in ℝ\}.$$

Fig. Examples of points in ${ℝ}^3$

The absolute value on ${ℝ}^n$ is the function $$\begin{array}{rrcl}\left|\phantom{x}\right|:& {ℝ}^n& ⟶& {ℝ}_{\ge 0}\\ & x& ⟼& \left|x\right|\end{array}\text{given by}\left|x\right|=\sqrt{x_1^2+\cdots +x_n^2}$$ for $x=(x_1,\dots ,x_n)$.

For example:

• If $n=2$ then $\left|\right(x,y\left)\right|=\sqrt{x^2+y^2}\in {ℝ}_{\ge 0}$.
• If $n=1$ then $\left|x\right|=\sqrt{x^2}\in {ℝ}_{\ge 0}$.

Lagrange's identity

(Lagrange's identity) $$\left(\sum _{i=1}^n{x}_i^2\right)\left(\sum _{i=1}^n{y}_i^2\right)-{\left(\sum _{i=1}^n{x}_i{y}_i\right)}^2=\frac{1}{2}\sum _{i,j=1}^n{(x_i{y}_j-x_j{y}_i)}^2.$$

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

For example, when $n=2$, OOPS THIS IS MESSED UP SOMEHOW $$\begin{array}{rcl}& & \frac{1}{2}({(x_1{y}_1-x_1{y}_1)}^2+{(x_1{y}_2-x_2{y}_1)}^2+{(x_2{y}_1-x_1{y}_2)}^2+{(x_2{y}_2-x_2{y}_2)}^2)\\ & =& {(x_1{y}_2-x_2{y}_1)}^2\\ & =& x_1^2{y}_2^2-2x_1{x}_2{y}_1{y}_2+x_2^2{y}_1^2\\ & =& (x_1^2+x_2^2)(y_1^2+y_2^2)-{(x_1{y}_1+x_2{y}_2)}^2.\end{array}$$

The inner product

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 $$⟨x,y⟩=(x_1,\dots x_n)\cdot \left(\begin{array}{c}{y}_1\\ ⋮\\ y_n\end{array}\right)=x_1{y}_1+\cdots +x_n{y}_n=\sum _{i=1}^n{x}_i{y}_i.$$

Note: The length of $x\in {ℝ}^n$ is given by $$\left|x\right|=\sqrt{x_1^2+\cdots +x_n^2}=\sqrt{⟨x,x⟩}.$$

(The Cauchy-Schwarz inequality) Let $x,y\in {ℝ}^n$. Then $$⟨x,y⟩\le \left|x\right|\left|y\right|.$$

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

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

		Proof.
	We have $$\begin{array}{rcl}{|x+y|}^2& =& ⟨x+y,x+y⟩\\ & =& ⟨x,x⟩+⟨x,y⟩+⟨y,x⟩+⟨y,y⟩& & =& {\left|x\right|}^2+2⟨x,y⟩+{\left|y\right|}^2\\ & \le & {\left|x\right|}^2+2\left|x\right|\left|y\right|+{\left|y\right|}^2\\ & =& {\left(\right|x|+|y\left|\right)}^2.\end{array}$$ So $$|x+y|\le \left|x\right|+\left|y\right|.\square$$ Notes and References This proof of the Cauchy-Schwarz and triangle inequality is that found in [Bou, ???]. A similar proof is found, in the complex case in [Ahl, ???]. An even better proof is to use the discriminant of the restriction of a nondegenerate form to the 2-dimensional subspace spanned by $x$ and $y$ as in [Bou, ???]. References [Bo] A. Borel, Linear Algebraic Groups, Section AG4.2, Graduate Texts in Mathematics 126, Springer-Verlag, Berlin, 1991, MR?????? [Go] R. Godement, Topologie algébrique et théorie des faisceaux, Section 1.9, Actualités scientifiques et industrielles 1252, Hermann, Paris, 1958. MR?????? page history