Cauchy-Bunyakovsky-Schwarz Inequality/Lebesgue 2-Space

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space.

Let $f, g: X \to \R$ be $\mu$-square integrable functions, that is $f, g \in \map {\LL^2} \mu$, Lebesgue $2$-space.

Then:


 * $\ds \int \size {f g} \rd \mu \le \norm f_2^2 \cdot \norm g_2^2$

where $\norm {\, \cdot \,}_2$ is the $2$-norm.

Equality
Equality in the above, that is:


 * $\ds \int \size {f g} \rd \mu \le \norm f_2^2 \cdot \norm g_2^2$

holds for almost all $x \in X$:
 * $\dfrac {\size {\map f x}^2} {\norm f_2^2} = \dfrac {\size {\map g x}^2} {\norm g_2^2}$

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

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