Theorem D.1.1. Cauchy-Schwarz.
For all \(x,y\in\R^n\text{,}\) \(|x\cdot y|\le \|x\|\,\|y\|\text{.}\)
Section D.1 Euclidean Norm on \(\R^n\)
For vectors \(x=(x_1,\dots,x_n)\) and \(y=(y_1,\dots,y_n)\) in \(\R^n\text{,}\) their dot product is \(x\cdot y=\sum_{i=1}^n x_i y_i\text{.}\) It is symmetric and bilinear.
The Euclidean norm of \(x\in\R^n\) is \(\|x\|=\sqrt{x\cdot x}\text{.}\) From the definition we immediately get \(\|x\|\ge 0\text{,}\) \(\|x\|=0\) if and only if \(x=0\text{,}\) and \(\|ax\|=|a|\,\|x\|\) for every \(a\in\R\text{.}\) When \(n=1\text{,}\) the norm is just the absolute value: \(\|x\|=|x|\text{.}\)
Proof.
If \(x=0\text{,}\) then both sides are zero, so there is nothing to prove. Assume \(x\ne0\text{.}\) For each \(t\in\R\text{,}\) \(\|y-tx\|^2\ge0\text{.}\) Expanding gives \(0\le \|y-tx\|^2
= (y-tx)\cdot(y-tx)
= \|y\|^2 - 2t(x\cdot y) + t^2\|x\|^2.\)
This quadratic polynomial in \(t\) is never negative, so its discriminant is at most zero: \(4(x\cdot y)^2 - 4\|x\|^2\|y\|^2 \le 0.\) Hence \((x\cdot y)^2\le\|x\|^2\|y\|^2\text{,}\) and taking square roots gives \(|x\cdot y|\le \|x\|\,\|y\|\text{.}\)
Proposition D.1.2.
For all \(x,y\in\R^n\text{,}\) \(\|x+y\|\le \|x\|+\|y\|\text{.}\)
Proof.
By bilinearity and Theorem D.1.1, \(\|x+y\|^2
= (x+y)\cdot(x+y)
= \|x\|^2 + 2(x\cdot y) + \|y\|^2
\le \|x\|^2 + 2\|x\|\,\|y\| + \|y\|^2
= (\|x\|+\|y\|)^2.\) Since both sides are nonnegative, taking square roots gives \(\|x+y\|\le \|x\|+\|y\|\text{.}\)
Proposition D.1.3.
For all \(x,y\in\R^n\text{,}\) \(|\|x\|-\|y\||\le \|x-y\|\text{.}\)
Proof.
By Proposition D.1.2, \(\|x\|=\|y+(x-y)\|\le\|y\|+\|x-y\|\text{,}\) so \(\|x\|-\|y\|\le\|x-y\|\text{.}\) Interchanging \(x\) and \(y\) gives \(\|y\|-\|x\|\le\|x-y\|\text{.}\) Therefore \(|\|x\|-\|y\||\le \|x-y\|\text{.}\)
