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

Theorem
Let $\mathbb K$ be a subfield of $\C$.

Let $V$ be a semi-inner product space over $\mathbb K$.

Let $x, y$ be vectors in $V$.

Then:
 * $\left\vert{\left \langle {x, y} \right \rangle}\right\vert^2 \le \left \langle {x, x} \right \rangle \left \langle {y, y} \right \rangle$

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 $\left \langle {y, y} \right \rangle \ne 0$.

Insert $\lambda = \left \langle {x, y} \right \rangle \left \langle {y, y} \right \rangle^{-1}$ in the inequality:

Reorder the inequality to get:


 * $\left\vert{\left \langle {x, y} \right \rangle}\right\vert^2 \le \left \langle {x, x} \right \rangle \left \langle {y, y} \right \rangle$

Next, suppose $\left \langle {y, y} \right \rangle = 0$.

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

Insert $\lambda = n \left \langle {x, y} \right \rangle$ in the inequality:

Rearrange the inequality to get:


 * $\left \langle {x, x} \right \rangle \ge 2n \left\vert{\left \langle {x, y} \right \rangle}\right\vert^2$

As this inequality holds for all $n \in \N$, it follows that $\left\vert{\left \langle {x, y} \right \rangle}\right\vert^2 = 0$.

Then:


 * $\left\vert{\left \langle {x, y} \right \rangle}\right\vert^2 \le 0 = \left \langle {x, x} \right \rangle \left \langle {y, y} \right \rangle$