Cauchy-Schwartz Inequality

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• From Euclidean Projection on a line, We have $|x^{T}y|\le ||x|{|}_2.||y|{|}_2$
• The angle, $\theta$ defined by $\cos \theta =\frac{x^{T}y}{||x|{|}_2.||y|{|}_2}$
• Implies that Cosine angle between 2 vector have magnitude less than one
• $\to$ for any $n$-vector $y:{\text{max}}_{||x|{|}_2\le 1}{x}^{T}y=||y|{|}_2$

Generalized Cauchy-Schwartz Inequality:

• For any two vectors $x,y\in {\mathbb{R}}^n:|x^{T}y|\le ||x|{|}_p||y|{|}_q$ where $\frac{1}{p}+\frac{1}{q}=1$
  • ex: $|x^{T}y|\le ||x|{|}_p||y|{|}_q$

Proof:

• WLOG, assume $||x|{|}_2=1$ and show that $|x^{T}y|\le ||y|{|}_2$ for every $y$
• For every $t\in \mathbb{R}:$
  • $0\le ||tx-y|{|}_2^2=t^2-2t(x^{T}y)+||y|{|}_2^2=(t-x^{T}y{)}^2+||y|{|}^2-(x^{T}y{)}^2$
• Pluggin $t=x^{T}y$ proves the result