Cauchy-Schwarz inequality

Cauchy-Schwarz inequality states that the inner product of two vectors is always less or equal to the product of the norms of the two vector in any inner product space.

Mathematical statement

Let $V$ be an inner product space over $\mathbb{F}$.

$$\forall u,v\in V,\; |\langle u,v\rangle|\le\|u\|\|v\|.$$

Furthermore,

$$|\langle u,v\rangle|=\|u\|\|v\|\iff \exists \lambda\in\mathbb{F},\; u=\lambda v$$

Proof

If $v=0$,

$$\begin{gathered} |\langle u,v\rangle| = 0, \|u\|\|v\| = 0 \\ \therefore |\langle u,v\rangle| = \|u\|\|v\| \end{gathered}$$

If $v\neq 0$,

Consider the orthogonal decomposition

$$u = \frac{\langle u, v \rangle}{\| v \|^2} v + u_2$$

where $u_2\bot v$.

By the generalised Pythagorean theorem,

$$\| u \|^2 = \left\| \frac{\langle u, v \rangle}{\| v \|^2}v \right\|^2 + \| u_2 \|^2 = \frac{|\langle u, v \rangle|^2}{\| v \|^2} + \| u_2 \|^2 $$

Since $v\neq 0$:

$$\begin{gathered} \frac{|\langle u, v \rangle|^2}{\| v \|^2} + \| u_2 \|^2 \ge \frac{|\langle u, v \rangle|^2}{\| v \|^2} \\ \Rightarrow \| u \|^2 \ge \frac{|\langle u, v \rangle|^2}{\| v \|^2} \\ \Rightarrow |\langle u,v\rangle|\le\|u\|\|v\|. \end{gathered}$$