Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces/Proof 1

Proof
Let $x, y \in V$.

Let $\lambda \in \mathbb K$.

Then:

where $\lambda^*$ denotes the complex conjugate of $\lambda$.

(If $\mathbb K$ is a subfield of $\R$, then $\lambda^* = \lambda$.)

First, suppose $\innerprod y y \ne 0$.

Insert $\lambda = \innerprod x y \innerprod y y^{-1}$ in the inequality:

Reorder the inequality to get:


 * $\size {\innerprod x y}^2 \le \innerprod x x \innerprod y y$

Next, suppose $\innerprod y y = 0$.

By Number Field has Rational Numbers as Subfield, $\Q \subset \mathbb K$.

Let $n \in \N \subset \Q \subset \mathbb K$ be a natural number.

Insert $\lambda = n \innerprod x y$ in the inequality:

Rearrange the inequality to get:


 * $\innerprod x x \ge 2 n \size {\innerprod x y}^2$

If $\size {\innerprod x y}^2 \ne 0$, then:
 * $\dfrac {\innerprod x x} {2 \size {\innerprod x y}^2} \ge n$

for every $n \in \N$, contradicting the Axiom of Archimedes.

Therefore, by Proof by Contradiction:
 * $\size {\innerprod x y}^2 = 0$.

Then:


 * $\size {\innerprod x y}^2 \le 0 = \innerprod x x \innerprod y y$