Cauchy-Bunyakovsky-Schwarz Inequality/Lebesgue 2-Space

Theorem
Let $\left({X, \Sigma, \mu}\right)$ be a measure space.

Let $f, g: X \to \R$ be $\mu$-square integrable functions, i.e. $f, g \in \mathcal{L}^2 \left({\mu}\right)$, Lebesgue $2$-space.

Then:


 * $\displaystyle \int \left\vert{f g}\right\vert \, \mathrm d \mu \le \left\Vert{f}\right\Vert_2^2 \cdot \left\Vert{g}\right\Vert_2^2$

where $\left\Vert{\cdot}\right\Vert_2$ is the $2$-norm.

Equality
Equality in the above, i.e.:


 * $\displaystyle \int \left\vert{f g}\right\vert \, \mathrm d \mu \le \left\Vert{f}\right\Vert_2^2 \cdot \left\Vert{g}\right\Vert_2^2$

holds iff for almost all $x \in X$: $\dfrac {\left\vert{f \left({x}\right)}\right\vert^2} {\left\Vert{f}\right\Vert_2^2} = \dfrac {\left\vert{g \left({x}\right)}\right\vert^2} {\left\Vert{g}\right\Vert_2^2}$

Proof
Follows directly from Hölder's Inequality with $p = q = 2$.

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