Some geometry: Cauchy-Schwarz inequality.

$\overrightarrow{x} ,\overrightarrow{y}\in \mathbb{R} ^3$	$f,g \in C[a,b]$
$\cos \angle \overrightarrow{x} ,\overrightarrow{y} =\frac {x_1y_1+x_2y_2+x_3y_3... ...+x_2^2+x_3^2} \cdot \sqrt{y_1^2+y_2^2+y_3^2}} \quad \overset{?}{\leq} 1 \qquad$	$\qquad \cos \angle f,g = \frac{\int_a^b f(x)g(x)dx} {\sqrt{\int_a^b f(x)^2dx} \cdot \sqrt{\int_a^b g(x)^2dx}} \quad \overset{?}{\leq} 1$

Question: Why is it possible to define $\cos \phi$ by the previous formulae? Answer: If the conditions for a function to be an inner product are fulfilled, then we can prove the Cauchy-Schwarz inequality:

Theorem 13.0.1   .

Corollary 13.0.2   $\forall \overrightarrow{x} ,\overrightarrow{y}\in V, \quad \left\vert \frac{\l... ...trix} \cdot \begin{Vmatrix}\overrightarrow{y}\end{Vmatrix}} \right\vert \leq 1$.