Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces

Theorem
Let $V$ be a semi-inner product space over $\mathbb K$ where $\mathbb K$ is a subfield of $\C$.

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 $\lambda \in \mathbb K$. Since an inner product is generated by a norm on the underlying normed linear space we may expand as follows:

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

(If $\mathbb K = \R$, then $\lambda^* = \lambda$.)

When $y = 0$ the equality is the trivial statement that $0 \leq 0$, so assume that $y \neq 0$.

If we let $\lambda = \left \langle {x, y} \right \rangle \times \left \langle {y, y} \right \rangle^{-1}$ then we obtain:


 * $0 \le \left \langle {x, x} \right \rangle - \left|{\left \langle {x, y} \right \rangle}\right|^2 \times \left \langle {y, y} \right \rangle^{-1}$

Solving this for $\left\vert{\left \langle {x, y} \right \rangle}\right\vert^2 $, we see that:


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

as desired.

Also known as
This theorem is also known as the Schwarz Inequality or Cauchy-Bunyakovsky-Schwarz Inequality.